Answer:
The angle θ between the direction of the current and the direction of the uniform magnetic field is 41° or 139°.
Explanation:
The force on a current carrying wire is given by the following equation:
[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]
The cross-product can be written with a sine term:
[tex]F = ILB\sin(\theta)\\0.025 = (3.4)(0.28)(0.04)\sin(\theta)\\\sin(\theta) = 0.6565[/tex]
Therefore, the angle θ is 41.03° and 138.97°
The angles can be calculated using the formula sin(θ) = F / (I L B), giving two symmetrical values about 90° in the first and second quadrants because the sine function is periodic.
The question is asking for the angles at which the force on a current-carrying wire in a magnetic field is a specific magnitude. The magnitude of the force exerted on a current-carrying wire placed in a magnetic field is given by the formula F = I L B sin(θ), where F is the force, I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.
By rearranging for θ, we get the equation sin(θ) = F / (I L B). Plugging in the values from the question, we find sin(θ) = 0.0250 N / (3.40 A imes 0.280 m imes 0.0400 T). This gives us θ values that correspond to the sine of this ratio.
There are two angles that will produce the same sine value because sine is a periodic function, which are θ and 180°-θ. Therefore, the two angles between the direction of the current and the direction of the uniform magnetic field for which the force on the wire has a magnitude of 0.0250 N will be symmetrical about 90° in the first and second quadrants.
A 12,000-lb bus collides with a 2800-lb car. The velocity of the bus before the collision is vB = 18i (ft/s) and the velocity of the car is vC =33j (ft/s). The two vehicles become entangled and remain together after the collision. The coefficient of kinetic friction between the vehicles’ tires and the road is μk = 0.6. Find (a) velocity of combined center of mass immediately after the collision, (b) final position (assume brakes are on and you have skidding, not rolling). (c) If the collision lasted 0.5 seconds, what was the impulsive force of the bus on the car?
Answer:
(a) 20.84 ft/s
(b) 11.24 ft
(c) -68160 N
Explanation:
Parameters given:
Mass of Bus, Mb = 12000 lb
Mass of car, Mc = 2800 lb
Initial speed of bus before collision, u(b) = 18 ft/s
Initial speed of car before collision, u(c) = 33 ft/s
Coefficient of friction, μk = 0.6
(a)Combined velocity after collision.
Since the bus and car are entangled and move together after the collision, they have the same velocity after collision.
Using the law of conservation of momentum, we have:
Mb*u(b) + Mc*u(c) = Mb*v(b) + Mc*v(c)
Where v(b) = velocity of the bus,L after collision,
v(c) = velocity of the car after collision.
Since v(b) = v(c) = v,
Mb*u(b) + Mc*u(c) = (Mb + Mc)*v
=> (12000 * 18) + (2800 * 33) = (12000 + 2800) * v
=> 308400 = 14800 * v
=> v = 20.84 ft/s
Combined velocity after collision is 20.84 ft/s
(b) The force acting on the bus and car after collision is given as:
F = m*a
Where F = force,
m = combined mass of bus and car,
a = acceleration of the bus and car after collision.
We know that the only force acting on the combined mass of the bus and car is the Frictional force, Fr and it is given as:
Fr = μk*m*g
Where g = acceleration due to gravity
Hence,
Fr = ma
=> - μk*m*g = m*a
The negative sign signifies that the fictional force is acting in the opposite direction to the lotion.
=> a = - μk*g
a = - 0.6 * 32.2
a = - 19.32 ft/s^2
Using one of the equations of linear motion, we can find the distance moved by the car and bus after collision:
v*v = u*u + 2*a*s
Where s = distance moved
The final velocity, v, of the car and bus is 0 because they come to rest and the initial velocity, u, is 20.84 ft/s, the velocity of the car and bus after collision.
Hence,
0 = (20.84*20.84) + (-2*19.32*s)
=> s = -434.3056/-38.64
s = 11.24 ft
(c) Impulsive force is the force that two bodies which are colliding exert on one another. It is given mathematically as
I. F. = (momentum change)/time
Momentum change of the bus is:
Momentum change = final momentum - initial momentum
Momentum change = Mb*v(b) - Mb*u(b)
Momentum change = (12000*20.84) - (12000*18)
Momentum change = 250080 - 216000
Momentum change = - 34080 kgft/s
=> I. F. = - 34080/0.5
I. F. = - 68160 N
The Impulsive force of the bus on the car is - 68160N
The negative sign means the force is acting opposite to the motion of the car.
Consider a concentric tube heat exchanger. Assuming there is no fin attached to any of the surfaces and considering negligible fouling, what would be the appropriate equation for calculating the overall heat transfer coefficient?
Answer:
[tex]Q = T1 - T2 (KA)/L[/tex]
Explanation:
Where Q is the co-efficient of heat transfer.
T-1 is initial temperature of exchanger
T-2 is final temperature of sink where has to to be dissipated
k is the co-efficient of thermal conductivity
A is the total area of exchanger surface...
and L is the total length of exchanger...
The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer? Select the best answer from the choices provided.
a) 1/6 seconds
b) 1/4 seconds
c) 1/2 seconds
d)1/3 seconds
e) 1/5 seconds
Answer:
1/3
Explanation:
Take the displacement and divide it by the speed of light. Meters will cancel leaving you with seconds. :)
110m / 343m/s = 0.32069971 seconds or 1/3 seconds
The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer 1/3 second ( approx.). So, option (d) is correct.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
So, distance between the first dancer in the line and the last dancer in the line be = 120 m -10 m = 110 m
Velocity of sound in air be = 343 m/s.
Hence, The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer = distance/velocity = 110/343 second = 1/3 second ( approx.).
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The distance from the earth to the sun is about 1.50×1011 m . Find the total power radiated by the sun.
Answer:
Power, [tex]P=3.93\times 10^{26}\ W[/tex]
Explanation:
Given that,
The distance from the earth to the sun is about, [tex]d=1.5\times 10^{11}\ m[/tex]
let us assume that the average intensity of solar radiation at the upper atmosphere of earth,. [tex]I=1390\ W/m^2[/tex]
We need to find the total power radiated by the sun. The intensity is defined as the total power divided by the area of a sphere of radius equal to the average distance between the earth and the sun. It is given by :
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
P is total power
[tex]P=4\pi r^2\times I[/tex]
[tex]P=3.93\times 10^{26}\ W[/tex]
So, the total power radiated by the sun is [tex]P=3.93\times 10^{26}\ W[/tex]. Hence, this is the required solution.
Final answer:
The total power radiated by the Sun is calculated using the area of the Sun and the power radiated per square meter at its surface. The Sun's total power output is found to be 3.82×1026 W. This value is important for understanding the energy Earth receives from the Sun, known as the solar constant (1360 W/m²).
Explanation:
The distance from the Earth to the Sun is about 1.50×1011 meters. The total power radiated by the Sun can be determined by using the following physics principle: The Sun, behaving as a perfect black body with an emissivity of exactly 1, radiates power uniformly across its surface area. The power radiated per square meter on the Sun's surface is found to be 6.3×107 W/m². The sun's radius is approximately 7.00×108 meters. Using the formula Power = Area × Intensity, we calculate the Sun's total power output.
The area of a sphere is given by 4πR2, where R is the radius of the sphere. Therefore, the total power output of the Sun is 4πR2σT4, which calculates to 3.82×1026 W. This immense amount of power is what we refer to as the solar luminosity.
At the distance of the Earth, this power is spread over a spherical area with a radius equal to the Earth-Sun distance. We can find the power per square meter at this distance, known as the solar constant, which is approximately 1360 W/m².
A power supply maintains a potential difference of 52.3 V 52.3 V across a 1570 Ω 1570 Ω resistor. What is the current in the resistor?
Answer:
0.033 A
Explanation:
Current: This can be defined as the rate of flow of electric charge in a circuit.
The S.I unit of current is Ampere (A)
From Ohm's law.
V = IR ............................ Equation 1
Where V = Potential difference, I = current, R = resistance.
Making I the subject of the equation,
I = V/R................... Equation 2
Given: V = 52.3 V, R = 1570 Ω
Substitute into equation 2
I = 52.3/1570
I = 0.033 A.
Hence the current in the resistor = 0.033 A
An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose instead it were already moving at speed v and the same amount of work W was done on it. What would be its final speed
Answer:
Explanation:
Given
W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)
According to work energy theorem work done by all the forces is equal to change in kinetic energy of object
[tex]W=\frac{1}{2}mv^2---1[/tex]
where m=mass of object
v=velocity of object
When the object is already have velocity v then the final speed is given by work energy theorem
[tex]W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2[/tex]
From 1 and 2 we get
[tex]\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2[/tex]
[tex]2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2[/tex]
[tex]v_f^2=2v^2[/tex]
[tex]v_f=\sqrt{2}v[/tex]
An isolated, parallel‑plate capacitor carries a charge Q . If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:a. Unchanged.b. Halved.c. Doubled.d. Quartered.e. Quadrupled.
Answer:
option C
Explanation:
Given,
Change on the capacitor = Q
Separation is doubled
Energy stored in the capacitor,E = ?
we know,
[tex]E = \dfrac{Q^2}{2C}[/tex]
and [tex]C = \dfrac{\epsilon_0A}{d}[/tex]
now,
[tex]E = \dfrac{Q^2}{2\epsilon A}\ d[/tex].......(1)
where d is the separation between the two plates.
now, when the separation is doubled
[tex]E' = \dfrac{Q^2}{2\epsilon A}\ (2d)[/tex]
[tex]E' = 2(\dfrac{Q^2}{2\epsilon_0 A}\ d)[/tex]
From equation (1)
E' = 2 E
Hence, the energy stored in the capacitor is doubled if the separation is increased.
The correct answer is option C.
"A boat that can travel at 4.0 km/h in still water crosses a river with a current of 2.0 km/h. At what angle must the boat be pointed upstream (that is, relative to its actual path) to go straight across the river?
Answer:
30 degrees
Explanation:
The boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.
What is vector law of addition ?
Vector addition is governed by the triangle law which states that when two vectors are represented by two triangle sides with their order of magnitude and direction, the resultant vector's magnitude and direction will be represented by the third triangle side.
It is given that speed of boat in still water v₁ = 4 km/h
speed of current v₂ = 2 km/h
Let the relative speed of boat with respect to current be = v km/h
and the angle made by v₁ with v be = θ
No find the angle θ with which the boat to go straight across the river uptream by triangle law of vector addition as shown below :
[tex]\begin{aligned}\theta &=\text{sin}^{-1}\left ( \frac{2}{4} \right )\\&= 30^{0}\end{aligned}[/tex]
Therefore, the boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.
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An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 970 kgkg traveling initially at a speed of 52.0 km/hkm/h in a distance equal to the diameter of a dime, which is 1.8 cmcm ?
Answer:
F=5618278.8 N
Explanation:
Given that
m = 970 kg
Initial speed ,u = 52 km/h
u = 14.44 m/s ( 1 km/h = 0.277 m/s)
Distance d= 1.8 cm = 0.018 m
The final speed ,v = 0 m/s
We know that
v² = u² + 2 a d
a=Acceleration
0² = 14.44² + 2 x a x 0.018
[tex]a=-\dfrac{14.44^2}{2\times 0.018}\ m/s^2[/tex]
a=5792.04 m/s²
We know that
Force = Mass x acceleration
F= m a
F= - 5792.04 x 970 N
F= - 5618278.8 N
Therefore the magnitude of the force will be 5618278.8 N.
F=5618278.8 N
Which of the following statements about insulating materials is correct?a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.b.The electric field from a charged object is able to penetrate through an insulating material.c.Insulating materials exhibit a linear relationship between voltage and current.d.Silver and copper are good insulators.
Answer:
a.Insulators can be used to increase the amount of current that can flow through a resistor without increasing its temperature.
Explanation:
A conductor has low resistance, while an insulator has much higher resistance. Devices called resistors control amounts of resistance into an electrical circuits. Electric charges inside an insulator are bound to the individual atoms or molecules, not being able to move inside the material.
When you turn up the resistance, the electric current flowing through the circuit is reduced
Ohm's law:
V = I * R
R = V/I
An Ohmic conductor would have a linear relationship between the current and the voltage. With non-Ohmic conductors, the relationship is not linear. Most metals are good conductors example silver, copper etc.
How do astronomers set about looking for extrasolar planets?
Explanation:
Implicit techniques for the discovery of extra-solar planets are used by astronomers. Evidence from the radial velocity of a change in the rotation of the star indicates the presence of a planet. If, with reference to Earth, the inclination of the planet's orbit is viewed as an edge-on, the detection of light originating from the star throughout its planetary transit would then be a verification.
The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is sufficiently loud to cause permanent hearing damage if the operator doesn't wear ear protection. What are (a) the sound intensity and (b) the sound intensity level for a person watching from 50 m away?
Answer:
(a) [tex]I_{1}=3.2*10^{-3}W/m^{2}[/tex]
(b) [tex]\beta =95dB[/tex]
Explanation:
Given data
Distance r₁=50 m
Distance r₂=2 m
Intensity I₂=2.0 W/m²
To find
(a) The Sound Intensity I₁
(b) The Sound Intensity level β
Solution
For (a) the Sound Intensity I₁
[tex]\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}[/tex]
For (b) the Sound Intensity level β
The Sound Intensity level β is calculated as follow
[tex]\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB[/tex]
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when it crosses over onto a floor that has a coefficient of kinetic friction = 0.35. How far will the puck travel across the floor before it stops?
Answer:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
Explanation:
For this case we can use the second law of Newton given by:
[tex] \sum F = ma[/tex]
The friction force on this case is defined as :
[tex] F_f = \mu_k N = \mu_k mg [/tex]
Where N represent the normal force, [tex]\mu_k [/tex] the kinetic friction coeffient and a the acceleration.
For this case we can assume that the only force is the friction force and we have:
[tex] F_f = ma[/tex]
Replacing the friction force we got:
[tex] \mu_k mg = ma[/tex]
We can cancel the mass and we have:
[tex] a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}[/tex]
And now we can use the following kinematic formula in order to find the distance travelled:
[tex] v^2_f = v^2_i - 2ad[/tex]
Assuming the final velocity is 0 we can find the distance like this:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
The hockey puck will travel approximately 58.5 meters across the floor before it stops.
Explanation:To solve this problem, we will use the concepts of kinetic energy and the work-energy theorem.
Initially, the hockey puck has a kinetic energy of KE1 = 0.5*m*v^2 = 0.5*0.3 kg*(20 m/s)^2 =60 J
As it travels across the floor, the friction does work on it until it stops. The work done by the friction force is W = μ*m*g*d, where: μ is the coefficient of kinetic friction (0.35), m is the mass (0.3 kg), g is the gravitational constant (9.8 m/s^2), and d is the distance traveled.
The work done by the friction is equal to the initial kinetic energy (work-energy theorem), so we can solve for d: 60 J = 0.35 * 0.3 kg * 9.8 m/s^2 * d. Simplifying this equation gives d = 60 J / (0.35 * 0.3 kg * 9.8 m/s^2) = 58.5 m
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There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2.0 × 107 m/s in a direction opposite to the field. Its speed increases to 4.0 × 107 m/s over a distance of 1.2 cm. What is the magnitude of the electric field?
Answer:
Explanation:
Given
speed of Electron [tex]u=2\times 10^7\ m/s[/tex]
final speed of Electron [tex]v=4\times 10^7\ m/s[/tex]
distance traveled [tex]d=1.2\ cm[/tex]
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=Final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}[/tex]
[tex]a=5\times 10^{16}\ m/s^2[/tex]
acceleration is given by [tex]a=\frac{qE}{m}[/tex]
where q=charge of electron
m=mass of electron
E=electric Field strength
[tex]5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}[/tex]
[tex]E=248.3\ kN/C[/tex]
3 An empty hot tub has a mass of 320 kg. When filled, the tub holds 600 gallons of water (rho = 62.4 lbm/ft3). The local acceleration due to gravity is 32 ft/s2. Determine the total weight of the hot tub and water in pounds-force (lbf)
Answer:
Total weight of the hot tub and water is 5676.6 pounds-force
Explanation:
rho = 62.4lbm/ft^3 × 1ft^3/7.481gal = 8.34lbm/gal
Mass of water = rho × volume = 8.34lbm/gal × 600 gallons = 5004lbm = 5004×0.45359kg = 2269.8kg
Total mass of hot tub and water = 320kg + 2269.8kg = 2589.8kg
Local acceleration due to gravity = 32ft/s^2 = 32ft/s^2 × 1m/3.2808ft = 9.75m/s^2
Total weight of hot tub and water = 2589.8kg × 9.75m/s^2 = 25250.55N = 25250.55/4.4482 lbf = 5676.6 lbf
A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with the ground of 35o. The coefficient of static friction between the box and the board is 0.5 and the coefficient of kinetic friction is 0.3 . Will the box slide down the ramp when it is at 35o?
Answer:
Yes it will slide down the ramp
Explanation:
Let m be the mass of the box and gravitational acceleration g = 9.81m/s2, we can calculate the gravity that affects the box
W = mg
When the box is at 35 degree incline, this gravity is split into 2 components, 1 parallel to the incline (Wsin35) and the other one perpendicular with the incline (Wcos35).
The one perpendicular with the incline has an equal and opposite normal force of Wcos35
This normal force would dictate the static friction force where coefficient = 0.5. So the static friction is 0.5mgcos35
The box would slide when the parallel component of gravity wins over static friction force
mgsin35 > 0.5mgcos35
Since mg is positive we can cancel them out on both sides
sin35 > 0.5cos35
0.57 > 0.5*0.82
0.57 > 0.41
This is true so we can conclude that the box slides down the ramp
A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
a) What is the magnitude of the force of the wall on ladder?b) What is the magnitude of the normal force of the ground on ladder?c) What is the minimum coefficient of friction so the ladder does not slip?d) What is the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder?
A) Force of the wall on the ladder: 186.3 N
B) Normal force of the ground on the ladder: 725.2 N
C) Minimum value of the coefficient of friction: 0.257
D) Minimum absolute value of the coefficient of friction: 0.332
Explanation:
a)
The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:
[tex]W=mg[/tex]: weight of the ladder, with m = 20 kg (mass) and [tex]g=9.8 m/s^2[/tex] (acceleration of gravity)
[tex]W_M=Mg[/tex]: weight of the person, with M = 54 kg (mass)
[tex]N_1[/tex]: normal reaction exerted by the wall on the ladder
[tex]N_2[/tex]: normal reaction exerted by the floor on the ladder
[tex]F_f = \mu N_2[/tex]: force of friction between the floor and the ladder, with [tex]\mu[/tex] (coefficient of friction)
Also we have:
L = 4.1 m (length of the ladder)
d = 3.0 m (distance of the man from point A)
Taking the equilibrium of moments about point A:
[tex]W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}[/tex]
where
[tex]Wsin 21^{\circ}[/tex] is the component of the weight of the ladder perpendicular to the ladder
[tex]W_M sin 21^{\circ}[/tex] is the component of the weight of the man perpendicular to the ladder
[tex]N_1 sin 69^{\circ}[/tex] is the component of the normal force perpendicular to the ladder
And solving for [tex]N_1[/tex], we find the force exerted by the wall on the ladder:
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N[/tex]
B)
Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of [tex]N_2[/tex].
We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.
Therefore, we have:
[tex]\sum F_y = 0\\N_2 - W - W_M =0[/tex]
And substituting and solving for N2, we find:
[tex]N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N[/tex]
C)
Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.
The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.
Therefore, we can write:
[tex]\sum F_x = 0\\F_f - N_1 = 0[/tex]
And re-writing the equation,
[tex]\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257[/tex]
So, the minimum value of the coefficient of friction is 0.257.
D)
Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.
From part C), we saw that the coefficient of friction can be written as
[tex]\mu = \frac{N_1}{N_2}[/tex]
This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}[/tex]
We see that this quantity is maximum when d is maximum, so when
d = L
Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)[/tex]
And substituting, we get
[tex]N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N[/tex]
And therefore, the minimum coefficient of friction in order for the ladder not to slip is
[tex]\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332[/tex]
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A solid sphere and a hollow sphere have the same mass and radius. The two spheres are spun with matching angular velocities. Which statement is true? I. The hollow has the greater angular momentum II. The solid sphere has the greater angular momentum. III. The angular momentum is the same for both spheres. IV. The moment of inertia is the same for both spheres.
Answer:
I. The hollow sphere has the greater angular momentum.
Explanation:
Given that the mass and radius of hollow sphere and solid sphere are same. Let the mass and radius of two spheres be m and r respectively. The two spheres are rotating having same angular velocity ω .
Moment of inertia of solid sphere, I₁ = [tex]\frac{2}{5}\times{m}r^{2}[/tex]
Moment of inertia of hollow sphere, I₂ = [tex]\frac{2}{3}\times{m}r^{2}[/tex]
Since, I₁ and I₂ are not equal. Therefore, the statement iv is wrong.
The relation between angular momentum, moment of inertia and angular velocity is :
L = Iω
Let L₁ and L₂ be the angular momentum of solid sphere and hollow sphere respectively.
L₁ = I₁ω and L₂ = I₂ω
As ω is same for both spheres but I₂ is greater than I₁, hence L₂ is greater than L₁.
Therefore, statement I is correct that the hollow sphere has the greater angular momentum.
The angular momentum of the hollow sphere is greater than that of solid sphere.
The moment of inertia of solid sphere is given as follows;
[tex]I_{ss} = \frac{2}{5} mr^2 = 0.4mr^2[/tex]
The moment of inertia of hollow sphere is given as follows;
[tex]I_{hs} = \frac{2}{3} mr^2 = 0.67 mr^2[/tex]
The angular momentum of each sphere is calculated as follows;
[tex]L = I\omega \\\\L_{ss} = 0.4mr^2 \omega \\\\L_{hs} = 0.67 mr^2 \omega[/tex]
Thus, we can conclude that the angular momentum of the hollow sphere is greater than that of solid sphere.
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What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizontal?
Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:
[tex]W_{work}=2.67*10^{-3}J[/tex]
Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
[tex]W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J[/tex]
A ball is thrown vertically upward with a speed of 19.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started?
Answer:
Explanation:
initial speed, u = 19 m/s
(a) Let it rises upto height h.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity and it is zero.
0 = 19 x 19 - 2 x 9.8 x h
h = 18.4 m
(c) Let the ball takes time t to reach to the maximum height.
use first equation of motion
v = u - gt
0 = 19 - 9.8 x t
t = 1.94 s
(c) The time taken by the ball to reach to the ground = 2 x time to reach to maximum height
T = 2 x t = 2 x 1.94 = 3.88 s
(d) When the ball reaches the ground, let the velocity is v.
Use third equation of motion
v² = u² - 2 gh
where, v is the final velocity
v² = 0 + 2 x 9.8 x 18.4
v = 19 m/s
(a) The ball reaches a height of approximately 18.68 meters. (b) It takes approximately 1.94 seconds to reach its highest point (c) The time taken to hit the ground is 1.94 seconds. When the ball returns to the level from which it started, its velocity is -19.0 m/s.
Explanation:(a) To find the height that the ball reaches, we can use the kinematic equation:
Δy = v2 / (2g)
where Δy is the change in height, v is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have:
Δy = (19.0 m/s)2 / (2 * 9.8 m/s2)
Calculating, we find that the ball rises to a height of approximately 18.68 meters.
(b) The time it takes for the ball to reach its highest point can be calculated using the equation:
t = v / g
where t is the time, v is the initial velocity, and g is the acceleration due to gravity. Substituting the given values, we get:
t = (19.0 m/s) / (9.8 m/s2)
Simplifying, we find that it takes approximately 1.94 seconds for the ball to reach its highest point.
(c) Since the time it takes for the ball to reach its highest point is the same as the time it takes for it to fall back down, the time it takes for the ball to hit the ground after reaching its highest point is also 1.94 seconds.
(d) When the ball returns to the level from which it started, its velocity is equal in magnitude but opposite in direction to its initial velocity. Therefore, the velocity when it returns is -19.0 m/s.
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A mass is attached to a spring with spring constant k = 20 N m . The spring is stretched to 10 cm past its resting position. How much work (in J) does the spring do when the object is released and the mass travels back to its initial position?
The spring does 0.1 Joules of work when the object is released and travels back to its initial position.
We have,
The work done by the spring can be calculated using the formula for the potential energy stored in a spring:
Potential Energy (PE) = 0.5 * k * x²,
where k is the spring constant and x is the displacement from the equilibrium position.
Given:
Spring constant (k) = 20 N/m,
Displacement (x) = 0.10 m (10 cm).
Calculate the potential energy when the spring is stretched to 10 cm:
PE = 0.5 * k * x²
= 0.5 * 20 N/m * (0.10 m)²
= 0.5 * 20 * 0.01 J
= 0.1 J.
The spring stores 0.1 Joules of potential energy when stretched to 10 cm.
When the mass is released and returns to its initial position, this potential energy is converted back into kinetic energy as the mass accelerates.
Therefore,
The spring does 0.1 Joules of work when the object is released and travels back to its initial position.
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Final answer:
The spring does 0.1 Joules of work when the mass travels back to its initial position.
Explanation:
To find the work done by a spring when an object is released and travels back to its initial position, we can use the formula:
Work = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant is 20 N/m and the displacement is 10 cm (which is 0.1 m). Plugging these values into the formula, we get:
Work = (1/2) * 20 N/m * (0.1 m)^2 = 0.1 J
So, the spring does 0.1 Joules of work when the mass travels back to its initial position.
A new, previously unknown, planet Vulcan was discovered in our solar system. We measure an orbital period of 103 Earth days for the new planet. We send some astronauts to travel to the planet and they measure the planet's gravitational acceleration to be 8.2 m/s2 on its surface. They also determine that the planet's radius is half the radius of Earth.
Q1. How far away from the sun is the new planet?
(A) 8.3 x 100 m
(B) 6.4 x 100 m
(C) 1.7 x 100 m
(D) 9.5 x 1010 m
(E) 2.3 x 1010 m
Q2. What is the mass of planet Vulcan?
(A) 1.3 x 1024 kg
(B) 5.4 x 1024 kg
(C) 9.8 x 1024 kg
(D) 0.2 x 102 kg
(E) 4.8 x 1024 kg
Q3. The astronauts also discover a moon that orbits planet Vulcan with a period of 63 days. How far away is the moon from the planet Vulcan?
(A) 1 x 108 m
(B) 20 x 108 m
(C) 460 x 108 m
(D) 9 x108 m
(E) 4 x 10 m
Q4. The astronauts visit the newly discovered moon to study it. They measure the gravitational acceleration on the surface of the moon to be 2.7 m/s. When their mission is finished, they try to escape from the moon. They measure that the minimum velocity to escape the moon is 3,000 m/s. What is the mass of the moon?
(A) 1.1 x 1023 kg
(B) 5.0 x 1023 kg
(C) 7.8 x 1023 kg
(D) 29 x 104 kg
(E) 370 x 1023 kg
Answer:
Q1. corect is B, Q2. it is A, Q3. E and Q4. A
Explanation:
Q1 For this exercise we can use Newton's second law where acceleration is centripetal.
F = m a
a = v² / r.
G m M / r² = m v² / r
G M / r = v²
The velocity has a constant magnitude whereby we can divide the length of the circular orbit (2π r) between the period
G M / r = (2π r / T)²
r³ = G M T2 / 4π²
Let's calculate
T = 103 day (24 h / 1 day) (3600 s / 1h) = 8,899 10⁶ s
r³ = 6.67 10⁻¹¹ 1.99 10³⁰ (8,899 10⁶) 2 / 4π²
r = ∛ (266.25 10³⁰)
r = 6.4 10¹⁰ m
The distance matches the value in part B
Q2 Astronauts have measured the acceleration of gravity, so we can use the second law with a body on the planet's surface
F = m g
G m M_p / R_p² = m g
G M_p / R_p² = g
M_p = g R_p² / G
They indicate that the radius of the planet is half the radius of the Earth
R_p = ½ R_earth
R_p = ½ 6.37 10⁶
R_p = 3.185 10⁶ m
Let's calculate
M_p = 8.2 (3,185 10⁶)² / 6.67 10⁻¹¹
M_p = 1.25 10²⁴ kg
The correct answer is A
Q3 We use Newton's second law again, with part Q1, where M is the mass of the planet and m is the mass of the moon
r³ = G M T² / 4π²
T = 63 days (24h / 1day) (3600s / 1h) = 5.443 10⁶ s
r³ = 6.67 10⁻¹¹ 1.25 10²⁴ (5.443 10⁶)² / 4π²
r = ∛ (62.56807 10²⁴)
r = 3.97 10⁸ m
The correct answer is E
Q4 To calculate this part let's use the conservation of mechanical energy,
Starting point The surface of the moon
Em₀ = K + U = ½ m v2 - G m M / r
Final point. Infinity with zero speed
[tex]Em_{f}[/tex] = 0
Em₀ = Em_{f}
½ m v² - G m M / R = 0
v² = 2 G M / r
M = v2 r / 2G
r = 2 G M / v²
Since we don't know the radius of the moon, we will also use the equation in part 2
M = g r² / G
r = √ GM / g
Let's replace
2G M / v² = √ G M / g
4 G M / v⁴ = 1 / g
M = v⁴ / (g 4G)
M = 3000⁴ / (2.7 4 6.67 10-11)
M = 1.12 10²³ kg
corract is A
The destination airport has one runway, 08-26, and the wind is calm. The normal approach in calm wind is a left hand pattern to runway 08. There is no other traffic at the airport. A thunderstorm about 6 miles west is beginning to develop to its mature stage, and rain is starting to reach the ground. The pilot decides to a. Fly normal approach to runway 8b. Fly a right hand approach to runway 8c. Fly the approach to runway 26d. Fly to the west for a fun ride
Answer:
C
Explanation:
Correct answer: C. Fly the approach to runway 26. The winds from the storm could suddenly gust up and you don’t want to be in a short final or even in the flare when a tailwind from a storm gusts up. No other traffic and calm winds currently means that you can land on any runway you want. Go for the safe bet and land on 26.
In this scenario, the pilot should fly a left hand approach to runway 08.
Explanation:In this scenario, with a calm wind, the pilot should fly a left hand approach to runway 08.
Since the wind is calm, there is no need for the pilot to adjust the approach. Flying in a clockwise pattern to runway 08 is the normal procedure.
Choosing any other option would not be appropriate or necessary.
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Assume that a clay model of a lion has a mass of 0.225 kg and travels on the ice at a speed of 0.85 m/s. It hits another clay model, which is initially motionless and has a mass of 0.37 kg. Both being soft clay, they naturally stick together.
What is their final velocity?
Answer:
Final velocity will be equal to 0.321 m/sec
Explanation:
We have given mass of clay model of lion [tex]m_1=0.225kg[/tex]
Its speed is 0.85 m/sec, so [tex]v_1=0.85m/sec[/tex]
Mass of another clay model [tex]m_2=0.37kg[/tex]
It is given that second clay is motionless
So its velocity [tex]v_2=0m/sec[/tex]
Now according to conservation of momentum
Momentum before collision will be equal to momentum after collision
So [tex]m_1v_!+m_2v_2=(m_1+m_2)v[/tex], here v is velocity after collision
So [tex]0.225\times 0.85+0.37\times 0+(0.225+0.37)v[/tex]
[tex]0.595v=0.191[/tex]
v = 0.321 m/sec
So final velocity will be equal to 0.321 m/sec
The net electric charge of an amber rod which has been rubbed with fur is called negative Group of answer choices because amber is an insulator by arbitrary convention so that the proton charge will be positive because like charges repel None of the above
Answer:
The right option is option E. None of the answer choices given are totally correct.
Explanation:
All insulators normally have an equal amount of positive and negative charges distributed on their surface.
The amber rod (an insulator) is called negative because after the coming together with fur (another insulator), the amber rod rubs off electrons from the fur onto itself and has an overall more negatively charged particles than positively charged particles on its surface.
The fur in turn becomes positive because it has more positive charges than negative on its surface.
So, the convention allows the now rubbed off amber rod to be called negative.
So, it is evident that none of the answer choices are totally correct, the right answer is more of a mix of some of the answer choices and more!
Hope this helps!!
A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s?
Answer:
Frequency of the vibration due to crevice is 1500 Hz.
Explanation:
Frequency is defined as rate of vibration per second. Its S.I. unit is s⁻¹ or Hz.
According to the problem, we have to find number of crevice car makes in 1 second.
Speed of car, u = 30 m/s
Distance covered by car in 1 second, d = 30 m
For every 0.02 m, one crevice occurred by the tire of car.
Number of crevice occurred in 1 second by the car = [tex]\frac{30}{0.02}[/tex]
= 1500
Since, each crevice makes a single vibration. Thus, the frequency of these vibrations is 1500 Hz.
The frequency of the vibrations generated by the tire treads in this case, given that each crevice is 2.00 cm apart and the car moves at 30.0 m/s, is 1500 Hz.
Explanation:In this scenario, it is important to understand that the frequency of the vibrations is determined by the speed of the tire and the tread pattern. Here, the crevice, which causes the vibration, occurs every 2.00 cm (or 0.02 m). This means for every meter the car moves, 0.02 m into 1 m gives a total of 50 vibrations. Therefore, at a speed of 30.0 m/s, we will have 50 vibrations/m multiplied by 30 m/s, giving a frequency of 1500 Hz or vibrations per second.
Frequency in this question refers to the number of times the vibration from the crevices occur in a unit of time (in this case, per second). This concept is crucial in understanding wave motions and vibration patterns, and is often applied in physics-related disciplines.
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the position of a crate sliding down a ramp is given by x=0.25t^3 m, z = 6-0.75t^5/2 m, where t is in seconds. determine the magnitude of the crates velocity and acceleration when t = 2 seconds
Answer:
v = 30.15 m/s
a = 60.07 m/s2
Explanation:
Velocity is derivative of position with respect to time
[tex]v_x = x' = 3*0.25t^2 = 0.75t^2[/tex]
[tex]v_z = z' = 5*0.75t^4/2 = 1.875t^4[/tex]
Acceleration is derivative of velocity with respect to time
[tex]a_x = v_x' = 2*0.75 t = 1.5t[/tex]
[tex]a_z = v_z' = 4*1.875t^3 = 7.5t^3[/tex]
At t = 2 seconds
[tex]v_x = 0.75*2^2 = 3m/s[/tex]
[tex]v_z = 1.875*2^4 = 30m/s[/tex]
[tex]v = \sqrt{v_x^2 + v_z^2} = \sqrt{3^2 + 30^2} = \sqrt{909} = 30.15 m/s[/tex]
[tex]a_x = 1.5*2 = 3 m/s^2[/tex]
[tex]a_z = 7.5*2^3 = 60 m/s^2[/tex]
[tex]a = \sqrt{a_x^2 + a_z^2} = \sqrt{3^2 + 60^2} = \sqrt{3609} = 60.07 m/s[/tex]
A vehicle of mass 1,500 kg is traveling at a speed of 50 km/hr. What is the kinetic energy stored in its mass? Calculate the energy that can be recovered by slowing the vehicle to a speed of 10 km/hr.
Answer:
Explanation:
Given
mass of vehicle [tex]m=1500\ kg[/tex]
Speed of vehicle [tex]u=50\ km/hr\approx 13.89\ m/s[/tex]
Kinetic Energy Possessed by mass
[tex]K_i=\frac{1}{2}mu^2[/tex]
[tex]K_i=\frac{1}{2}\times 1500\times (13.89)^2[/tex]
[tex]K_i=144.69\ kJ[/tex]
when vehicle is slowed down to speed of [tex]v=10\ km/hr\approx 2.78\ m/s[/tex]
Kinetic Energy at this speed
[tex]K_f=\frac{1}{2}mv^2[/tex]
[tex]K_f=\frac{1}{2}\times 1500\times (2.78)^2[/tex]
[tex]K_f=5.78\ kJ[/tex]
Energy Recovered [tex]=K_i-K_f[/tex]
Energy Recovered[tex]=144.69-5.78=138.9\ kJ[/tex]
A friend is writing a science fiction screenplay about an asteroid on a collision course with Earth. She asks you to calculate some numbers so her scenario will be correct. Astronauts will attach a rocket engine to the asteroid in an attempt to divert it. The asteroid is moving at 21 km/s. The rocket will provide an acceleration of 0.035 km/s2 at a right angle to the original motion. The rocket only has enough fuel to provide this acceleration for 40 seconds. Will this change the direction of the asteroid’s motion by at least 22°, enough to miss Earth and save civilization? (15 pts, according to Grading for problem solving, see reverse side. Your group should submit one analysis, with all group member’s names, either on the Problem solving framework or the plain white paper provided.)
Answer:
No, the deviation in the path of asteroid is not by 22°
Explanation:
Given:
velocity of asteroid, [tex]v_a=21\ km.s^{-1}[/tex]
acceleration of the rocket, [tex]a=0.035\ km.s^{-2}[/tex]
time of acceleration, [tex]t=40\ s[/tex]
Now, the final velocity of the asteroid:
using the equation of motion,
[tex]v=u+a.t[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity in the direction
[tex]v=0+0.035\times 40[/tex]
[tex]v=1.4\ km.s^{-1}[/tex]
Now direction of the resultant velocity:
[tex]\tan\beta=\frac{v_a}{v}[/tex]
[tex]\tan\beta=\frac{21}{1.4}[/tex]
[tex]\beta=86.186^{\circ}[/tex]
So, the deviation in the asteroid:
[tex]\theta=90-\beta[/tex]
[tex]\theta=90-86.186[/tex]
[tex]\theta=3.814^{\circ}[/tex]
What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?
Answer:
0.044 V
Explanation:
E = Electric field = [tex]5.5\times 10^6\ V/m[/tex]
d = Thickness of membrane = 8 nm
When the electric field strength is multiplied by the membrane thickness we get the voltage
Voltage across a gap is given by
[tex]V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V[/tex]
The voltage across the membrane is 0.044 V
Final answer:
The voltage across an 8.00 nm-thick membrane with an electric field strength of 5.50 MV/m is calculated using the formula V = Ed, resulting in a voltage of 44.0 mV.
Explanation:
The voltage across a membrane can be determined by using the relationship between electric field strength (E), voltage (V), and the distance (d) the electric field spans. The electric field is uniform and the formula to use is V = Ed. In this case, the electric field (E) is given as 5.50 MV/m or 5.50 x 10⁶V/m, and the thickness of the membrane (d) is 8.00 nm or 8.00 x 10⁻⁹ m.
To find the voltage across the membrane, we simply multiply the electric field by the thickness of the membrane:
V = (5.50 x 10⁶ V/m) x (8.00 x 10⁻⁹ m)
Therefore:
V = 44 x 10⁻³ V
V = 44.0 mV
So, the voltage across an 8.00 nm-thick membrane with the given electric field strength is 44.0 mV.