Answer:
a) [tex](x_2,y_2)=(-45.6,58.8)[/tex]
b) [tex]s=74.4097\ m[/tex]
Explanation:
Given:
x-component of avg. velocity, [tex]v_x=-3.8\ m.s^{-1}[/tex]x-component of avg. velocity, [tex]v_y=4.9\ m.s^{-1}[/tex]initial position of rhino, [tex](x_1,y_1)=(0,0)[/tex]initial time, [tex]t_1=0\ s[/tex]final time, [tex]t_2=12\ s[/tex]a)
As we know that average velocity is total displacement per unit time.
Position of x-coordinate of rhino:
[tex]v_x=\frac{x_2}{t_2}[/tex]
[tex]-3.8=\frac{x_2}{12}[/tex]
[tex]x_2=-45.6\ m[/tex]
Position of y-coordinate of rhino:
[tex]v_y=\frac{y_2}{t_2}[/tex]
[tex]4.9=\frac{y_2}{12}[/tex]
[tex]y_2=58.8\ m[/tex]
b)
Now the distance from the origin:
[tex]s=\sqrt{x_2^2+y_2^2}[/tex]
[tex]s=\sqrt{(-45.6)^2+(58.8)^2}[/tex]
[tex]s=74.4097\ m[/tex]
The question concerns vector kinematics in physics. The rhinoceros's coordinates at 12 seconds are (-45.6,58.8), and its distance from the origin is approximately 74.2 meters.
Explanation:The phenomenon described in your question essentially relates to Vector Kinematics. The rhinoceros is said to have a velocity in the x-direction of -3.8 m/s and in the y-direction of 4.9 m/s. These velocities are constant and last for 12 seconds.
To find the x- and y-coordinates, we can simply multiply the component velocities by time.
For the x-coordinate:
x = velocity_x * time = -3.8 m/s * 12 s = -45.6 m
And for the y-coordinate:
y = velocity_y * time = 4.9 m/s * 12 s = 58.8 m
So, for part (a) of your question, the rhino's coordinates at t2 = 12.0 s are (-45.6, 58.8).
Now, for part (b), to calculate the rhino's distance from the origin, we can use the Pythagorean theorem which is rooted in the geometrical interpretation of vectors. The distance from the origin (r) can be given by:
r = sqrt[x² + y²] = sqrt[(-45.6 m)² + (58.8 m)²] = approximately 74.2 m
Therefore, the rhino is approximately 74.2 m from the origin.
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A bullet moving with an initial speed of vo strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true of the collision? o The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. the bullet immediately after the collision the potential energy of the bullet and block at the instant they reach the maximum O The initial momentum of the bullet before the collision is equal to the momentum of o The kinetic energy of the bullet and block immediately after the collision is equal to height h. energy of the bullet and block when they reach the maximum heighth energy of the bullet and block immediately after the colision O The initial kinetic energy of the bullet before the collision is equal to the potential o The initial kinetic energy of the bullet before the collision is equal to the kinetic
Final answer:
The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. Option A
Explanation:
The correct statement of the collision in this scenario is A) The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system. In this case, the bullet and block form a closed system, and therefore, the initial momentum of the bullet is equal to the momentum of the bullet and block at the maximum height.
To understand why this statement is true, we can consider the momentum of the bullet and block system at different points in the motion. Initially, the bullet has momentum in the positive x-direction, and this momentum is transferred to the bullet and block as they rise to the maximum height. Therefore, the statement A is correct.
Suppose a hydrogen atom in its ground state moves 130 cm through and perpendicular to a vertical magnetic field that has a magnetic field gradient dB/dz = 1.2 × 102 T/m. (a) What is the magnitude of the force exerted by the field gradient on the atom due to the magnetic moment of the atom's electron, which we take to be 1 Bohr magneton? (b) What is the vertical displacement of the atom in the 130 cm of travel if its speed is 2.2 × 105 m/s?
Answer:a)1.11×10^-21Nm
b) 1.16×10^-3m
Explanation:see attachment
A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radius 2RR that also carries charge QQ. The charge QQ is distributed uniformly over the insulating shell.
A. Find the magnitude of the electric field in the region 02R. Express your answer in terms of the variables R, r, Q, and constants
π
and
ε
0.
Answer:
[tex]E=0[/tex] at r < R;
[tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex] at 2R > r > R;
[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex] at r >= 2R
Explanation:
Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,
[tex]\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}[/tex] (integral over a closed surface)
where,
[tex]E[/tex] = Electric field
[tex]Q_{enclosed}[/tex] = charged enclosed within the closed surface
[tex]\epsilon[/tex] = permittivity of free space
Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .
at r < R,
[tex]Q_{enclosed}[/tex] = 0 and hence [tex]E[/tex] = 0 (since the sphere is conducting, all the charges get repelled towards the surface)
at 2R > r > R,
[tex]Q_{enclosed}[/tex] = Q,
therefore,
[tex]E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}[/tex]
(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to [tex]4\pi r^{2}[/tex])
or, [tex]E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}[/tex]
at r >= 2R
[tex]Q_{enclosed}[/tex] = 2Q
Hence, by similar calculations, we get,
[tex]E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}[/tex]
The electric field inside the solid conducting sphere is zero. For r between R and 2R, the electric field can be calculated using Gauss's law.
Explanation:The electric field inside a solid conducting sphere is zero. Thus, the magnitude of the electric field in the region 0<r<R is zero.
In the region R<r<2R, the charge on the insulating shell induces an equal and opposite charge on the inner surface of the conducting sphere. Therefore, the magnitude of the electric field in this region can be found using Gauss's Law.
The electric field magnitude is given by:
E = Q / (4πε0 r2)
where Q is the charge on the insulating shell, ε0 is the vacuum permittivity, and r is the distance from the center of the sphere.
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A small boat is moving at a velocity of 3.35m/s when it is accelerated by a river current perpendicular to the initial direction of motion. The current accelerates the boat at 0.750m/s^2. what will the new velocity (magnitude and direction) of the boat be after 5 s?
Final answer:
To calculate the new velocity of the boat after being accelerated by the current for 5 seconds, one must use the Pythagorean theorem and arctangent function to find the magnitude and direction of the resultant velocity, resulting in approximately 5.04 m/s at 48.1 degrees from the original motion.
Explanation:
The student's question is about calculating the resultant velocity of a boat being accelerated by a river current perpendicular to its initial direction of motion. With an initial velocity of 3.35 m/s and a river current accelerating the boat at 0.750 m/s2, we need to find the new velocity after 5 seconds.
First, calculate the velocity increase caused by the acceleration of the current: velocity increase = acceleration × time, which gives us 0.750 m/s2 × 5 s = 3.75 m/s. This increase is perpendicular to the initial velocity.
Now, we determine the resultant velocity using the Pythagorean theorem since the velocities are perpendicular. The magnitude of the resultant velocity, Vtotal, is given by √(3.35 m/s)2 + (3.75 m/s)2, which equals to approximately 5.04 m/s. To find the direction, we use the arctangent function: θ = tan-1(3.75 m/s / 3.35 m/s), which results in a direction of approximately 48.1 degrees relative to the original direction of motion. Therefore, the boat will be moving with a velocity of approximately 5.04 m/s at a direction of approximately 48.1 degrees from its original direction, due to the current.
Two balls of different radii, 53 cm and 26 cm move directly toward each other with the same speed. If they are originally at a center-to-center distance 223 cm and it takes them 18.9 s to collide, how fast were the balls moving?
Answer:
[tex]v = 3.81\ m/s[/tex]
Explanation:
given,
Radius of the ball, r₁ = 53 cm
Radius of another ball, r₂ = 26 cm
center to center distance between the balls = 223 cm
time, t = 18.9 s
surface to surface distance between them
S = 223 - (53+26)
S = 144 cm
Speed of the ball = ?
[tex]relative\ speed = \dfrac{distance}{time}[/tex]
[tex]2 v = \dfrac{144}{18.9}[/tex]
both the balls are moving towards each other so, speed doubles.
[tex]v= \dfrac{7.62}{2}[/tex]
[tex]v = 3.81\ m/s[/tex]
Speed of the balls is equal to 3.81 m/s
Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 6.00 m/s. The velocity of the ball relative to Mia is 5.00 m/s in a direction 30.0 east of south. What are the magnitude and direction of the velocity of the ball relative to the ground?
Answer:
v_b = 10.628 m /s (13.605 degrees east of south)
Explanation:
Given:
- Velocity of mia v_m = + 6 j m/s
- Velocity of ball wrt mia v_bm = 5.0 m/s 30 degree due east of south
Find:
What are the magnitude and direction of the velocity of the ball relative to the ground? v_b
Solution:
- The relation of velocity in two different frame is given:
v_b - v_m = v_bm
- Components along the direction of v_b,m:
v_b*cos(Q) - v_m*cos(30) = 5
v_b*cos(Q) = 5 + 6 sqrt(3) / 2
v_b*cos(Q) = 5 + 3sqrt(3)
- Components orthogonal the direction of v_b,m:
-v_m*sin(30) = v_b*sin(Q)
-6*0.5 = v_b*sin(Q)
-3 = v_b*sin(Q)
- Divide two equations:
tan(Q) = - 3 / 5 + 3sqrt(3)
Q = arctan(- 3 / 5 + 3sqrt(3)
Q = -16.395 degrees
v_b = -3 / sin(-16.395)
v_b = 10.63 m/s
The magnitude of the soccer ball's velocity relative to the ground is 2.89 m/s, and its direction is approximately 56.31° north of east. This result is found by breaking down vector components and applying vector addition.
Explanation:You're asking about the velocity of the ball relative to the ground when soccer players Mia and Alice are running, and Alice passes the ball to Mia. To solve this, we'll use vector addition. Mia is running due north at 6.00 m/s, and the velocity of the ball relative to Mia is 5.00 m/s at 30° east of south. To find the velocity of the ball relative to the ground, we imagine two vectors: Mia's velocity vector (northward) and the ball's velocity vector relative to Mia. The latter will have an east and a south component due to the 30° angle.
To find the south component of the ball's relative velocity, we use cosine because it's adjacent to the 30° angle:
5.00 m/s * cos(30°) = 4.33 m/s. The east component is found using sine:
5.00 m/s * sin(30°) = 2.50 m/s.
Since Mia is running north, to find the actual velocity of the ball to the south, we subtract Mia's velocity from the south component:
4.33 m/s - 6.00 m/s = -1.67 m/s, where the negative indicates that the ball's actual movement is to the north.
Now, using the Pythagorean theorem for the ball's velocity relative to the ground, we find the magnitude:
(2.50 m/s)² + (-1.67 m/s)² = √(6.25 + 2.7889)
√8.3389 m²/s² = 2.89 m/s.
To find the direction, we use the tangent function, since we have opposite (east component) and adjacent (north component) sides of the right triangle:
an(θ) = 2.50 / 1.67; θ = tan(⁻¹)(2.50 / 1.67);
θ ≈ 56.31° north of east, which is the direction of the ball's velocity relative to the ground.
A manometer containing oil (rho = 850 kg/m3) is attached to a tank filled with air. If the oil-level difference between the two columns is 110 cm and the atmospheric pressure is 98 kPa, determine the absolute pressure of the air in the tank.
Answer:
[tex]P_{abs}=107172.35Pa\\ P_{abs}=107.172kPa[/tex]
Explanation:
Given data
[tex]P_{a}=98kPa\\ p_{oil}=850kg/m^{3}\\ h=110cm=1.10m[/tex]
To find absolute pressure of air in the tank.We must find the pressure athe parallel point from the right tube
So
[tex]P_{abs}=P_{a}+hp_{oil}g\\P_{abs}=98000Pa+(1.10m*850kg/m^{3}*9.81 )\\P_{abs}=107172.35Pa\\ P_{abs}=107.172kPa[/tex]
Which of the following statements is/are true? Check all that apply. Check all that apply. The total mechanical energy of a system is constant only if nondissipative interactions occur. Mechanical energy can be dissipated to nonmechanical forms of energy. The total mechanical energy of a system is equally divided between kinetic and potential energy. The total mechanical energy of a system is constant only if dissipative interactions occur. The total mechanical energy of a system, at any one instant, is either all kinetic or all potential energy.
Answer:
1) True, 2) True, 3) False, 4) False, 5) False
Explanation:
1) True. Dissipative energy cannot be recovered, in general it is a form of heat
2) True. The dissipation can be by radiation, heat
3) False. Mechanical energy is divided into K and U but not in equal parts
4) False. When there are dissipative interactions, part of the mechanical energy is set in the form of heat, so its value decreases
5) False. Mechanical energy is the sum of those two energies
The correct statements are a. The total mechanical energy of a system is constant only if nondissipative interactions occur. and b.Mechanical energy can be dissipated to non-mechanical forms of energy. Mechanical energy is conserved only if no dissipative interactions occur, and it can be transformed into nonmechanical forms. Mechanical energy isn't always equally divided between kinetic and potential energy and isn't necessarily completely one or the other at any instant.
Let's analyze the provided statements about mechanical energy:
The total mechanical energy of a system is constant only if nondissipative interactions occur. This statement is true. Mechanical energy remains constant in a system where only conservative forces act, meaning no dissipative (non-conservative) forces like friction are present.Mechanical energy can be dissipated to nonmechanical forms of energy. This statement is also true. Mechanical energy can be transformed into other forms like thermal energy due to dissipative forces such as friction.The total mechanical energy of a system is equally divided between kinetic and potential energy. This statement is not necessarily true; mechanical energy can vary between kinetic and potential energy and isn't always equally divided.The total mechanical energy of a system is constant only if dissipative interactions occur. This statement is false. Mechanical energy is only conserved if no dissipative forces act on the system.The total mechanical energy of a system, at any one instant, is either all kinetic or all potential energy. This statement is false. Mechanical energy is generally a sum of both kinetic and potential energy at any given instant and can vary in proportion.Thus, the correct statements are a. The total mechanical energy of a system is constant only if nondissipative interactions occur. and b.Mechanical energy can be dissipated to non-mechanical forms of energy.
How many revolutions per minute would a 26 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the topmost point?
Answer:[tex]N=8.28\ rpm[/tex]
Explanation:
Given
Diameter of wheel [tex]d=26\ m[/tex]
Person is feeling Weightlessness i.e. Net force on the person is equivalent to its weight
At top point weight is equal to Centripetal force
[tex]mg=\frac{mv^2}{r}[/tex]
where v=velocity of wheel
thus
[tex]g=\frac{v^2}{R}[/tex]
[tex]v=\sqrt{gR}[/tex]
[tex]v=\sqrt{9.8\times 13}[/tex]
[tex]v=11.28\ m/s[/tex]
[tex]v=\frac{\pi d\cdot N}{60}[/tex]
[tex]11.28=\frac{\pi \cdot 26\cdot N}{60}[/tex]
[tex]N=8.28\ rpm[/tex]
Answer:
Explanation:
Let m be the mass of passenger.
diameter of wheel, d = 26 m
radius of wheel, r = half of diameter = 13 m
Let ω be the angular velocity of the Ferris wheel.
A the passengers becomes weightless, so the centripetal force acting on the passengers is balanced by the weight of passengers.
mg = m r ω²
9.8 = 12 x ω²
ω = 0.9 rad/s
Let f be the frequency
ω = 2 π f
0.9 = 2 x 3.14 x f
f = 0.143 revolutions per second
Number of revolutions per minute = 0.143 x 60
= 8.6 revolutions per minute
Two equal positive point charges are placed at two of the co ?rners of an equilateral triangle of side A. What is the magnitude of the net electric field at the center of the triangle ?
Answer:
Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C
Explanation:
Given,A = side of the triangle. q = 1 C
The center of a triangle is the centroid of the triangle.
The distance between centroid to any vertices of a equilateral triangle is
[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle
[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]
[tex]=\frac{1}{\sqrt{3} } A[/tex]
Electric field= [tex]\frac{Kq}{d^2}[/tex] K=8.99×10⁹ Nm²/C², q=charge and d = distance
Therefore the magnitude of the net electric filed at the center of the triangle is
=2[tex]\frac{Kq}{d^2}[/tex] [both charge are at same distance from the centroid]
=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]
[tex]=\frac{6K}{A^2}[/tex] N /C [K=8.99×10⁹ Nm²/C²]
A 3.0 L cylinder is heated from an initial temperature of 273 K at a pressure of 105 kPa to a final temperature of 381 K. 381 K. Assuming the amount of gas and the volume remain the same, what is the pressure (in kilopascals) of the cylinder after being heated?
Answer:
[tex]{P_2}=146.53\ kPa[/tex]
Explanation:
Volume ,V = 3 L
Initial temperature ,T₁ = 273 K
Initial pressure ,P₁ = 105 kPa
Final temperature ,T₂ = 381 K
Lets take final pressure =P₂
We know that ,If the volume of the gas is constant ,then we can say that
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]
[tex]{P_2}=P_1\times \dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]{P_2}=105\times \dfrac{381}{273}\ kPa[/tex]
[tex]{P_2}=146.53\ kPa[/tex]
Therefore the final pressure will be 146.53 kPa.
The final pressure of the cylinder after being heated is 75.57 kPa.
Explanation:To solve this problem, we can use the equation for Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature. In this case, the initial volume is 3.0 L and the initial temperature is 273 K. The final temperature is 381 K. Now we can set up a proportion:
(Initial volume) / (Initial temperature) = (Final volume) / (Final temperature)
Plugging in the numbers, we get:
(3.0 L) / (273 K) = (Final volume) / (381 K)
Solving for the final volume gives us:
Final volume = [(3.0 L) / (273 K)] x (381 K) = 4.1732 L
Since the volume remains the same, the pressure is inversely proportional to the new volume. So, if the initial pressure is 105 kPa, the final pressure can be calculated using the following equation:
(Initial pressure) x (Initial volume) = (Final pressure) x (Final volume)
Plugging in the numbers, we get:
(105 kPa) x (3.0 L) = (Final pressure) x (4.1732 L)
Solving for the final pressure gives us:
Final pressure = [(105 kPa) x (3.0 L)] / (4.1732 L) = 75.57 kPa
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A particle of mass 73 g and charge 67 µC is released from rest when it is 47 cm from a second particle of charge −25 µC. Determine the magnitude of the initial acceleration of the 73 g particle. Answer in units of m/s 2 .
Answer:
933.804423995 m/s²
Explanation:
[tex]q_1[/tex] = Charge on particle 1 = 67 µC
[tex]q_2[/tex] = Charge on particle 2 = -25 µC
r = Distance between the particles = 47 cm
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
m = Mass of particle = 73 g
Electric force is given by
[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times -25\times 10^{-6}\times 67\times 10^{-6}}{0.47^2}\\\Rightarrow F=-68.1677229516\ N[/tex]
The magnitude of force is 68.1677229516 N
Acceleration is given by
[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{68.1677229516}{0.073}\\\Rightarrow a=933.804423995\ m/s^2[/tex]
The acceleration is 933.804423995 m/s²
A battery with an emf of 24.0 V is connected to a resistive load. If the terminal voltage of the battery is 16.1 V and the current through the load is 3.90 A, what is the internal resistance of the battery?
Answer:
2.03 Ω
Explanation:
EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.
The formula of emf is given as,
E = I(R+r)............................ Equation 1
Where E = Emf, I = current, R = External resistance, r = internal resistance.
Make r the subject of the equation
r = (E-IR)/I........................ Equation 2
Note: From ohm's law, V = IR.
r = (E-V)/I........................ Equation 3
Where V = Terminal voltage
Given: E = 24 V, I = 3.9 A, V = 16.1 V.
Substitute into equation 3
r = (24-16.1)/3.9
r = 7.9/3.9
r = 2.03 Ω
Final answer:
To find the internal resistance of a battery, subtract the terminal voltage from the emf and divide by the current. Given an emf of 24.0 V, a terminal voltage of 16.1 V, and a current of 3.90 A, the internal resistance is about 2.03 Ω.
Explanation:
To determine the internal resistance of a battery, we need to understand a critical relationship between the battery's electromotive force (emf), its terminal voltage, the current flowing through the circuit, and its internal resistance. The formula required for finding the internal resistance is derived from Ohm's Law, and it takes into account the difference between the emf (the voltage the battery would supply in the absence of internal resistance) and the terminal voltage (the actual voltage the battery supplies when connected to a load).
Given are the following parameters:
emf (E) = 24.0 V
Terminal Voltage (V) = 16.1 V
Current (I) = 3.90 A
To calculate the internal resistance (r), we use the formula based on the definition of emf, which is:
E = V + Ir
By rearranging the formula to solve for the internal resistance, we get:
r = (E - V) / I
Substituting the given values:
r = (24.0 V - 16.1 V) / 3.90 A = 7.9 V / 3.90 A ≈ 2.03 Ω
Therefore, the internal resistance of the battery is approximately 2.03 Ω.
A vinyl record makes 90 rotations in a minute. The diameter of the disk is 34 cm. Find the linear velocity of a point on the circumference of the disk in m/s.
Answer: 1.6 m/s
Explanation:
The relationship between linear speed and angular speed for a rotational motion is
[tex]v=\omega r[/tex]
where
v is the linear speed
[tex]\omega[/tex] is the angular speed
r is the distance from the axis of rotation
For the vinyl record here, we have:
[tex]\omega=90 rev/min[/tex]
Keeping in mind that
[tex]1rev = 2\pi rad\\1 min =60 s[/tex]
We can convert it to rad/s:
[tex]\omega = 90 \cdot \frac{2\pi}{60}=9.4 rad/s[/tex]
The diameter of the disk is 34 cm, so the radius is
[tex]r=\frac{34}{2}=17 cm = 0.17 m[/tex]
Therefore, the linear velocity of a point on the circumference is
[tex]v=(9.4)(0.17)=1.6 m/s[/tex]
The linear velocity of a point on the circumference of the disk is approximately 1.45 meters per second.
To find the linear velocity of a point on the circumference of the disk, we need to calculate the circumference of the disk and then determine how fast a point on the edge of the disk moves in one minute.
[tex]\[ v = \frac{C}{t} \][/tex]
First, we calculate the circumference using the diameter of the disk. The radius is half of the diameter, so:
[tex]\[ r = \frac{diameter}{2} = \frac{34 \text{ cm}}{2} = 17 \text{ cm} \][/tex]
Now, the circumference is given by:
[tex]\[ C = 2\pi r = 2\pi \times 17 \text{ cm} \][/tex]
Converting centimeters to meters (1 meter = 100 centimeters), we get:
[tex]\[ C = 2\pi \times 0.17 \text{ m} \][/tex]
Next, we know that the record makes 90 rotations in a minute, so the time \( t \) for one rotation is:
[tex]\[ t = \frac{60 \text{ seconds}}{90 \text{ rotations}} \][/tex]
Now we can calculate the linear velocity \( v \):
[tex]\[ v = \frac{C}{t} = \frac{2\pi \times 0.17 \text{ m}}{\frac{60}{90} \text{ seconds}} \] \[ v = \frac{2\pi \times 0.17 \text{ m}}{\frac{2}{3} \text{ seconds}} \] \[ v = \frac{2\pi \times 0.17 \text{ m} \times 3}{2} \] \[ v = \pi \times 0.17 \text{ m} \times 3 \] \[ v = \pi \times 0.51 \text{ m} \] Using the approximation \( \pi \approx 3.14159 \), we get: \[ v \approx 3.14159 \times 0.51 \text{ m/s} \] \[ v \approx 1.594 \text{ m/s} \][/tex]
Rounding to two decimal places, the linear velocity is approximately 1.45 meters per second."
What is spectroscopy?
Final answer:
Spectroscopy is a technique used in physics to study the absorption, emission, or scattering of electromagnetic radiation by atoms or molecules.
Explanation:
Spectroscopy is a technique used in physics to study the absorption, emission, or scattering of electromagnetic radiation by atoms or molecules. It is commonly used to analyze and understand the composition, temperature, and motion of celestial objects like stars and galaxies. By observing the patterns of light emitted or absorbed by these objects, astronomers can determine the elements present and gain insights into their physical properties.
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows to a halt. What forces act on the sled just after she's jumped in?
a) Gravity and kinetic friction
b) Gravity and normal force
c) Gravity and the force of the push
d) Gravity, a normal force, and kinetic friction
e) Gravity, a normal force, kinetic friction, and the force of the push
Answer:
d) Gravity, a normal force, and kinetic friction
Explanation:
When the bobsledder pushes her sled across horizontal snow to get it going, after she jumps into the sled there acts a force of gravity on the total mass of the sled including the bobsledder.The sled moves horizontally and not vertically this means that there is a normal force acting in the vertically upward direction opposite to the gravity.While the sled moves on the horizontal surface and comes to the rest there acts a kinetic frictional force on the body in the direction opposite to the direction of motion.Calculate the linear momentum of photons of wavelength 740 nm. What speed does anelectron need to travel to have the same linear momentum?
Answer:
v = 9.824 x 10³ m/s
Explanation:
given,
Linear momentum of photon,λ = 740 n m
for photon,
[tex]p=\dfrac{h}{\lambda}[/tex]
h is the planks constant
P is the momentum
[tex]p=\dfrac{6.626\times 10^{-34}}{740\times 10^{-10}}[/tex]
p = 8.95 x 10⁻²⁷ kg.m/s
For electron
p = m v
mass of electron = 9.11 x 10⁻³¹ Kg
[tex]v = \dfrac{8.95\times 10^{-27}}{9.11\times 10^{-31}}[/tex]
v = 9.824 x 10³ m/s
hence, the velocity of electron is equal to v = 9.824 x 10³ m/s
Your cell phone works as both a radio transmitter and receiver. Say you receive a call at a frequency of 880.65 MHz. What is the wavelength in meters?
Answer:
0.34m
Explanation:
Final answer:
The wavelength of a radio signal received at a frequency of 880.65 MHz is approximately 0.34 meters, using the formula λ = c / f, where c is the speed of light.
Explanation:
If you receive a call at a frequency of 880.65 MHz, to calculate the wavelength in meters, you can use the formula λ = c / f, where λ is the wavelength in meters, c is the speed of light (approximately 3.00 × 108 m/s), and f is the frequency in hertz (Hz). Since the question provides the frequency in megahertz (MHz), we first convert it to hertz by multiplying by 106.
The frequency is 880.65 MHz, which is equal to 880.65 × 106 Hz. Thus, the wavelength λ = 3.00 × 108 m/s / (880.65 × 106 Hz).
Calculating this gives a wavelength of approximately 0.34 meters.
Suppose you walk 17.5 m straight west and then 22.0 m straight north. How far are you from your starting point (in m)
Answer:
Explanation:
Given
Man walks 17.5 m straight to west 17.5 m
So position vector is given by
[tex]\vec{r_1}=-17.5\hat{i}[/tex]
Now he walks 22 m North
so position vector is
[tex]r_{21}=22\hat{j}[/tex]
Position of man from initial Position
[tex]\vec{r_{2}}=\vec{r_2}-\vec{r_1}[/tex]
[tex]\vec{r_{2}}=22\hat{j}-(-17.5\hat{i})[/tex]
[tex]\vec{r_{2}}=17.5\hat{i}+22\hat{j}[/tex]
So Magnitude of distance is given by
[tex]|\vec{r_{2}}|=\sqrt{17.5^2+22^2}[/tex]
[tex]|\vec{r_{2}}|=28.11\ m[/tex]
To determine the distance from the starting point after walking 17.5 m west and 22.0 m north, use the Pythagorean theorem. Calculating this gives a distance of approximately 28.1 meters from the starting point.
Calculating Distance Using Pythagorean Theorem
To find how far you are from your starting point after walking 17.5 m west and then 22.0 m north, we can use the Pythagorean theorem. This is because your path forms a right triangle with the two legs being 17.5 m and 22.0 m.
The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the distance from your starting point) is equal to the sum of the squares of the other two sides:
a² + b² = c²
a = 17.5 m (west)b = 22.0 m (north)Substituting these values into the Pythagorean theorem gives:
17.52 + 22.02 = c²
Calculating the squares:
306.25 + 484.00 = c²
Adding them together:
790.25 = c²
Taking the square root of both sides:
c ≈ 28.1 m
Therefore, you are approximately 28.1 meters from your starting point.
A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0 s, what is the current in the wire? 4.5 A 18 A 0.22 A 9.0 A If the current is a constant 3.0 A, how long will it take for 14.0 C of charge to move past a particular spot in the wire? 0.21 s 4.7 s 14.0 s 42 s
1) Current: 4.5 A
2) Time taken: 4.7 s
Explanation:
1)
The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing a given point in a time t
For the wire in this problem, we have
q = 9.0 C is the amount of charge
t = 2.0 s is the time interval
Solving for I, we find the current:
[tex]I=\frac{9.0}{2.0}=4.5 A[/tex]
2)
To solve this problem, we can use again the same formula
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing a given point in a time t
In this problem, we have:
I = 3.0 A (current)
q = 14.0 C (charge)
Therefore, the time taken for the charge to move past a particular spot in the wire is
[tex]t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s[/tex]
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The electric current in the wire in the first scenario is 4.5 A, and it will take 4.7 seconds for 14.0 C of charge to move past a particular spot on the wire with a constant current of 3.0 A in the second scenario.
Explanation:The amount of electric current in a circuit is defined by the amount of electric charge that passes through it in a given amount of time. This is represented by the formula I = Q / t, where I is the current, Q is the charge and t is the time.
In the first part of your question, we are given that Q = 9.0 C and t = 2.0 s. We can substitute these values into the formula to find the current: I = 9.0 C / 2.0 s = 4.5 A.
In the second part, we are given that I = 3.0 A and Q = 14.0 C. This time, we need to rearrange the formula to find t: t = Q / I = 14.0 C / 3.0 A = 4.7 s.
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A light-rail commuter train speeds up at a rate of 1.65 m/s 2 and takes 19.0 s to reach its top speed. How far down the tracks does the train go to reach its top speed if starting from rest?
Answer:
297.8 m
Explanation:
We are given that
Acceleration=[tex]a=1.65m/s^2[/tex]
Time,t=19 s
We have to find the distance covered by the train when it reach its top speed if starting from rest.
Initial speed=u=0
[tex]s=ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]s=0(19)+\frac{1}{2}\times 1.65(19)^2[/tex]
[tex]s=297.8 m[/tex]
Hence, the train covered 297.8 m when the train go to reach its top speed if starting from rest.
A sperm whale can accelerate at about 0.0900 m/s 2 when swimming on the surface of the ocean. How far will a sperm whale travel if it starts at a speed of 0.700 m/s and accelerates to a speed of 2.18 m/s ? Assume the whale travels in a straight line.
Answer:
s = 23.68 m
Explanation:
given,
acceleration of the sperm whale = 0.09 m/s²
initial speed of the whale, u = 0.7 m/s
final speed of the whale, v = 2.18 m/s
Distance traveled by the sperm whale = ?
Using equation of motion
v² = u² + 2 a s
2.18² = 0.7² + 2 x 0.09 x s
0.18 s = 4.2624
s = 23.68 m
Distance traveled by the whale is equal to s = 23.68 m
Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz. What will be the increase in frequency if these waves are reflected from a 95.0mi/h fastball headed straight toward the gun?
Since the Units presented are not in the International System we will proceed to convert them. We know that,
[tex]1 mi/h = 0.447 m/s[/tex]
So the speed in SI would be
[tex]V=95mi/h(\frac{0.447m/s}{1mi/h})[/tex]
[tex]V=42.465 m/s[/tex]
The change in frequency when the wave is reflected is
[tex]f'=f(1+\frac{V}{c})[/tex]
Or we can rearrange the equation as
[tex]f' = f + f\frac{V}{c}[/tex]
f' = Apparent frequency
f = Original Frequency
c = Speed of light
[tex]f'-f = f\frac{V}{c}[/tex]
[tex]\Delta f = f\frac{V}{c}[/tex]
Replacing,
[tex]\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})[/tex]
[tex]\Delta f =1489.8 Hz[/tex]
Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency
[tex]\Delta f_T = 2 \Delta f[/tex]
[tex]\Delta f_T = 2(1489.8Hz)[/tex]
[tex]\Delta f_T = 2979.63Hz[/tex]
Therefore the increase in frequency is 2979.63Hz
The increase in frequency of these waves is equal to 2979.6 Hertz.
Given the following data:
Frequency = 10.525 GHz.Velocity = 95.0 mi/h.Conversion:
1 mi/h = 0.447 m/s.
95.0 mi/h = [tex]95 \times 0.447[/tex] = 42.465 m/s.
To determine the increase in frequency:
How to calculate the increase in frequency.Mathematically, the change in frequency of a wave is given by this formula:
[tex]F' = F(1+\frac{V}{c} )\\\\F' = F+F\frac{V}{c}\\\\F' - F=F\frac{V}{c}\\\\\Delta F = F\frac{V}{c}[/tex]
Where:
F is the observed frequency.[tex]F'[/tex] is the apparent frequency.c is the speed of light.V is the velocity of an object.Substituting the given parameters into the formula, we have;
[tex]\Delta F = 10.525 \times 10^9 \times \frac{42.465}{3 \times 10^8}\\\\\Delta F = \frac{446.944 \times 10^9}{3 \times 10^8}\\\\\Delta F = 1489.8\;Hertz[/tex]
Since the waves were reflected, the increase in frequency toward the gun is double (twice) the change in frequency. Thus, we have;
[tex]F_2 = 2\Delta F\\\\F_2 = 2\times 1489.8\\\\F_2 = 2979.6\;Hertz[/tex]
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(a) If θ is in standard position, then the reference angle θ is the acute angle formed by the terminal side of θ and the . So the reference angle for θ = 110° is θ = °, and that for θ = 210° is θ =
Answer:
70 degree, 30 degree
Explanation:
We are given that [tex]\theta[/tex] is in standard position.
The reference angle [tex]\theta[/tex] is the acute angle.
Acute angle :The angle is always less than 90 degree.
The reference angle formed by terminal side of angle [tex]\theta[/tex] with x-axis.
When [tex]\theta=110^{\circ}[/tex]
Then, the reference angle =[tex]\bar{\theta}=180-110[/tex]
[tex]\bar{\theta}=70^{\circ}[/tex]
When [tex]\theta=210^{\circ}[/tex]
Therefore, the reference angle
[tex]\bar{\theta}=210-180[/tex]
[tex]\bar{\theta}=30^{\circ}[/tex]
In mathematics, a reference angle is the acute angle formed by the terminal side of a given angle and the x-axis, used for angles in standard position. The reference angle for θ = 110° is 70° and for θ = 210° it is 30°.
Explanation:The concept of reference angles is often taught in mathematics, particularly in trigonometry to help understand the properties of angles in different quadrants of the Cartesian plane. A reference angle is always the acute angle (less than 90°) that the terminal side of the given angle makes with the x-axis. In standard position, the reference angle for an angle is the smallest angle between the terminal side of the angle and the horizontal axis (x-axis).
For an angle θ of 110°, which lies in the second quadrant, the reference angle would be 180° - 110° = 70°. Similarly, for an angle θ of 210° that lies in the third quadrant, the reference angle is calculated as 210° - 180° = 30°. Thus, the reference angle for θ = 110° is 70°, and for θ = 210°, it is 30°.
The value of q is and the value of r is 75.0 cm. Note that, in this question, you are only asked to find the magnitude of the net force, but you should also think about the direction of the net force. What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls
Answer:
F_net = - 0.365 N (Down-ward direction)
Explanation:
Given:
- Value of r = 0.75 m
- Charges on x axis are -2*q
- Charge +q on origin
- Charge on - y axis is -2q
- Charge on + y axis is +q
- q = 5.00 * 10^-6 C
Find:
-What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls?
Solution:
- Force due to each of the two charges on x axis:
F_x = k*(-2*q)*(+q) / r*^2
r* = sqrt(2)*r
F_x = -k*q^2 / r^2 (Down-wards)
- Force due to +q charge on origin:
F_y = k*(+q)*(+q) / r^2
F_y = + k*q^2 / r^2 (Up-wards)
- Force due to -2*q charge on y-axis:
F_-2y = k*(-2*q)*(+q) / 4*r^2
F_-2y = - k*q^2 / 2*r^2 (Downwards-wards)
- Total net Force on charge +q on + y-axis:
2*F_x*sin(45) + F_y + F_-2y = F_net
-sqrt(2)*k*q^2 / r^2 + k*q^2 / r^2 - k*q^2 / 2*r^2 = F_net
(0.5-sqrt(2))*k*q^2 / r^2 = F_net
F_net = (0.5-sqrt(2))*(8.99*10^9)*(5*10^-6)^2 / 0.75^2
F_net = - 0.365 N
A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium streetlamps (λ = 589 nm) on Earth to be invisible to the astronauts who look with a powerful telescope upon the city streets of Earth?
Answer:
doppler shift's formula for source and receiver moving away from each other:
λ'=λ°√(1+β/1-β)
Explanation:
acceleration of spaceship=α=29.4m/s²
wavelength of sodium lamp=λ°=589nm
as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm
using doppler shift's formula:
λ'=λ°√(1+β/1-β)
putting the values:
700nm=589nm√(1+β/1-β)
after simplifying:
β=0.17
by this we can say that speed at that time is: v=0.17c
to calculate velocity at an acceleration of a=29.4m/s²
we suppose that spaceship started from rest so,
v=v₀+at
where v₀=0
so v=at
as we want to calculate t so:-
t=v/a v=0.17c ,c=3x10⁸ ,a=29.4m/s²
putting values:
=0.17(3x10⁸m/s)/29.4m/s²
t=1.73x10⁶
In the compound MgS, the sulfide ion has 1. lost one electrons. 2. lost two electrons. 3. gained one electron. 4. gained two electrons.
Answer:
4.has gained two electrons
Explanation:
There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.
Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2
Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6
This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.
Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.
Final answer:
In MgS, the sulfide ion has gained two electrons, resulting in a charge of -2, represented as S²-. Magnesium loses two electrons to form Mg²+, balancing the electron transfer to create a stable ionic compound.
Explanation:
In the compound MgS, the sulfide ion has gained two electrons to achieve a stable electron configuration. When sulfur (S), which has an atomic number of 16, gains two electrons, it results in an ion with 18 electrons and 16 protons. This gives the ion a charge of -2, since there are two more negative electrons than positive protons. Therefore, the sulfide ion is represented as S²-. The magnesium atom loses its two valence electrons to become a Mg²+ cation, as magnesium is in Group 2A of the periodic table and tends to lose two electrons to achieve a noble gas electron configuration. This electron transfer process is balanced, meaning the number of electrons lost by magnesium is equal to the number of electrons gained by sulfur, forming a stable ionic compound.
A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster turns on, causing an acceleration of 0.310 m/s^2 in the +x direction. The acceleration lasts for 49.0 s, at which point the thruster turns off.
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm.
a. 40cm/s,
b. 160 cm/s2,
c. 32cm/s,
d. -96cm/s2,
e. 0.232s
Answer:
a) [tex]v_{max}=0.4\ m.s^{-1}[/tex]
b) [tex]a_{max}=1.6\ m.s^{-2}[/tex]
c) [tex]v_x=0.32\ m.s^{-1}[/tex]
d) [tex]a_x=0.96\ m.s^{-1}[/tex]
e) [tex]\Delta t=0.232\ s[/tex]
Explanation:
Given:
mass of the object attached to the spring, [tex]m=0.5\ kg[/tex]
spring constant of the given spring, [tex]k=8\ N.m^{-1}[/tex]
amplitude of vibration, [tex]A=0.1\ m[/tex]
a)
Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.
Max. spring potential energy:
[tex]PE_s=\frac{1}{2} .k.A^2[/tex]
[tex]PE_s=0.5\times 8\times 0.1^2[/tex]
[tex]PE_s=0.04\ J[/tex]
When this whole spring potential is converted into kinetic energy:
[tex]KE_{max}=0.04\ J[/tex]
[tex]\frac{1}{2}.m.v_{max}^2=0.04[/tex]
[tex]0.5\times 0.5\times v_{max}^2=0.04[/tex]
[tex]v_{max}=0.4\ m.s^{-1}[/tex]
b)
Max. Force of spring on the mass:
[tex]F_{max}=k.A[/tex]
[tex]F_{max}=8\times 0.1[/tex]
[tex]F_{max}=0.8\ N[/tex]
Now acceleration:
[tex]a_{max}=\frac{F_{max}}{m}[/tex]
[tex]a_{max}=\frac{0.8}{0.5}[/tex]
[tex]a_{max}=1.6\ m.s^{-2}[/tex]
c)
Kinetic energy when the displacement is, [tex]\Delta x=0.06\ m[/tex]:
[tex]KE_x=PE_s-PE_x[/tex]
[tex]\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2[/tex]
[tex]\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2[/tex]
[tex]v_x=0.32\ m.s^{-1}[/tex]
d)
Spring force on the mass at the given position, [tex]\Delta x=0.06\ m[/tex]:
[tex]F=k.\Delta x[/tex]
[tex]F=8\times 0.06[/tex]
[tex]F=0.48\ N[/tex]
therefore acceleration:
[tex]a_x=\frac{F}{m}[/tex]
[tex]a_x=\frac{0.48}{0.5}[/tex]
[tex]a_x=0.96\ m.s^{-1}[/tex]
e)
Frequency of oscillation:
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{8}{0.5} }[/tex]
[tex]\omega=4\ rad.s^{-1}[/tex]
So the wave equation is:
[tex]x=A.\sin\ (\omega.t)[/tex]
where x = position of the oscillating mass
put x=0
[tex]0=0.1\times \sin\ (4t)[/tex]
[tex]t=0\ s[/tex]
Now put x=0.08
[tex]0.08=0.1\times \sin\ (4t)[/tex]
[tex]t=0.232\ s[/tex]
So, the time taken in going from point x = 0 cm to x = 8 cm is:
[tex]\Delta t=0.232\ s[/tex]
Final answer:
The problem involves calculating the maximum speed, maximum acceleration, speed and acceleration at a certain distance from equilibrium, and the time interval for an object in simple harmonic motion. Using formulas for SHM including maximum speed (v_max = ωA), maximum acceleration (a_max = ω^2A), and expressions for speed and acceleration at a given position, one can determine these values for the object on the spring.
Explanation:
The student has asked us to calculate various properties of an object undergoing simple harmonic motion (SHM) when attached to a spring with a known force constant and amplitude. To answer this question, one needs to use equations that describe SHM.
Maximum Speed (v_max) Calculation:
The maximum speed (v_max) of an object in SHM occurs when it passes through the equilibrium point and can be calculated using the formula v_max = ωA, where ω is the angular frequency (ω = sqrt(k/m)) and A is the amplitude of the motion.
Maximum Acceleration (a_max) Calculation:
The maximum acceleration (a_max) occurs at the maximum displacement and is given by a_max = ω^2A.
Speed at 6 cm from Equilibrium:
To find the speed at a certain position x, we use the formula v = ω sqrt(A^2 - x^2).
Acceleration at 6 cm from Equilibrium:
Acceleration at any position x is a = -ω^2x.
Time Interval to Move from 0 to 8 cm:
The time interval to move from x = 0 to a certain position x can be found using the formula for time in SHM as a function of position.
If you are at the equator and driving north at a speed of 90 m/s, what is direction of the magnetic force on your head? 1. north 2. south 3. downward 4. east 5. upward 6. west 7. There is no force.
Answer:
7. The force is zero
Explanation:
The force is zero when your velocity is parallel to the magnetic field