Answer:
D) result of the calculation is zero
Explanation:
Coulomb's Law is valid for only point-like particles. Since the ring is not a point-like, then we have to choose an infinitesimal portion (ds) of the ring, apply the Coulomb's Law to this portion and then integrate over the ring to find the total force.
The small portion (dq) will have the same charge density as the ring itself. Furthermore, the length of the infinitesimal portion is equal to the radius times the corresponding angle, dθ.
[tex]\lambda = \frac{Q}{2\pi R} = \frac{dq}{Rd\theta}\\dq = \frac{Qd\theta}{2\pi}[/tex]
Therefore, the force between the charge at the center and the small portion is
[tex]dF = \frac{1}{4\pi\epsilon_0}\frac{qdq}{R^2} = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}[/tex]
Since force is a vector, we have separate its x- and y-components,
[tex]dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\cos(\theta)\\dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQd\theta}{2\pi R^2}\sin(\theta)[/tex]
Now, we can integrate both of them over the ring.
[tex]F_x = \int\limits^{2\pi}_0 dF_x = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\cos(\theta)d\theta = 0\\F_y = \int\limits^{2\pi}_0 dF_y = \frac{1}{4\pi\epsilon_0}\frac{qQ}{2\pi R^2}\int\limits^{2\pi}_0\sin(\theta)d\theta = 0[/tex]
Since the integration from 0 to 2π for sine and cosine functions results as zero.
Therefore, the force on the charge at the center of a uniformly distributed ring is equal to zero.
The total force exerted on a charge q placed at the center of a circle with a uniformly distributed charge Q along the circumference is zero due to the symmetry of the charge distribution.
Explanation:When a charge Q is spread uniformly along the circumference of a circle with radius R, and a point particle with charge q is placed at the center of this circle, we must apply Coulomb's law to calculate the force exerted on the charge q. Thanks to the symmetry of the charge distribution, the forces exerted by individual segments of the charged circumference on the central charge will cancel each other out in every direction. Hence, while the distance from the charge q to any point on the circle is R, the resulting total force on charge q will be zero due to symmetry.
It's important not to confuse the circumference with other distances, such as the diameter (2R) or the circumferential length (2πR), as these are not relevant for calculating the force on the central charge in this symmetric setup. Therefore, the correct answer is that the result of the calculation is zero (Option D), because the uniform distribution of charge Q around the circle results in an equilibrium of forces.
Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.46 kg mass to the spring’s other end. The acceleration of gravity is 9.81 m/s 2 . If the spring stretches 3.7 cm from its equilibrium position, what is the spring constant?
Answer: The value of spring constant is 121.9 N/m
Explanation:
Force is defined as the mass multiplied by the acceleration of the object.
[tex]F=m\times g[/tex]
where,
F = force exerted on the object = ?
m = mass of the object = 0.46 kg
g = acceleration due to gravity = [tex]9.81m/s^2[/tex]
Putting values in above equation, we get:
[tex]F=0.46kg\times 9.81m/s^2=4.51N[/tex]
To calculate the spring constant, we use the equation:
[tex]F=k\times x[/tex]
where,
F = force exerted on the spring = 4.51 N
k = spring constant = ?
x = length of the spring = 3.7 cm = 0.037 m (Conversion factor: 1 m = 100 cm)
Putting values in above equation, we get:
[tex]4.51N=k\times 0.037m\\\\k=\frac{4.51N}{0.037m}=121.9N/m[/tex]
Hence, the value of spring constant is 121.9 N/m
A metal wire 1.50 m long has a circular cross section of radius 0.32 mm and an end-to-end resistance of 90.0 Ohms. The metal wire is then stretched uniformly so that its cross-section is still circular but its total length is now 6.75 m. What is the resistance of the wire after stretching? (Units: Ohm.)
Answer:
So after stretching new resistance will be 0.1823 ohm
Explanation:
We have given initially length of the wire [tex]l_1=150m[/tex]
Radius of the wire [tex]r_1=0.32mm=0.32\times 10^{-3}m[/tex]
Resistance of the wire initially [tex]R_1=90ohm[/tex]
We know that resistance is equal to [tex]R=\frac{\rho l}{A}[/tex] ,here [tex]\rho[/tex] is resistivity, l is length and A is area
From the relation we can say that [tex]\frac{R_1}{R_2}=\frac{l_1}{l_2}\times \frac{A_2}{A_1}[/tex]
Now length of wire become 6.75 m
Volume will be constant
So [tex]A_1l_1=A_2l_2[/tex]
So [tex]\pi \times (0.32)^2\times150=\pi \times r_2^2\times 6.75[/tex]
[tex]r_2=1.508mm[/tex]
So [tex]\frac{90}{R_2}=\frac{150}{6.75}\times \frac{1.508^2}{0.32^2}[/tex]
[tex]R_2=0.1823ohm[/tex]
With the new length and cross-sectional area, we determine the new resistance to be approximately 1822.5 Ohms.
To determine the resistance of the wire after it is stretched, follow these steps:
Calculate the initial volume of the wire using the initial length and cross-sectional area.
The initial length (L1) = 1.50 m
The radius (r1) = 0.32 mm = 0.00032 m
Initial cross-sectional area (A1) = πr1² = π (0.00032 m)² = 3.216 × 10⁻⁷ m²
Initial volume (V) = A1 × L1 = 3.216 × 10⁻⁷ m² × 1.50 m = 4.824 × 10⁻⁷ m³
Since volume remains constant, calculate the new radius after stretching.
The new length (L2) = 6.75 m
Initial volume (V) = New volume (V)
V = A2 × L2; thus, A2 = V / L2 = 4.824 × 10⁻⁷ m³ / 6.75 m = 7.148 × 10⁻⁸ m²
New radius (r2) = √(A2 / π) = √(7.148 × 10⁻⁸ m² / π) ≈ 0.000151 m = 0.151 mm
Calculate the new resistance using the resistivity formula.
Resistance (R) = ρ × L / A
Assuming resistivity (ρ) is the same, R1 / R2 = (L1 / A1) / (L2 / A2)
New resistance (R2) = R1 × (L2 / L1)² = 90 Ω × (6.75 m / 1.50 m)²
R2 = 90 Ω × (4.5)² = 90 Ω × 20.25 ≈ 1822.5 Ω
Therefore, the resistance of the wire after stretching is approximately 1822.5 Ohms.
An object is moving along the x-axis. At t = 0 it has velocity v0x = 20.0 m/s. Starting at time t = 0 it has acceleration ax = - Ct, where C has units of m/s3. (a) What is the value of C if the object stops in 8.00 s after t = 0? (b) For the value of C calculated in part (a), how far does the object travel during the 8.00 s?
The value of C if the object stops in 8.00 s is 0.625 m/s³.
The distance traveled by the object before stopping in 8 seconds is 40 m.
The given parameters;
initial velocity, [tex]v_0[/tex] = 20.0 m/sinitial time of motion, t = 0acceleration of the object, a = -CtThe value of C is determined by using velocity equation as shown below;
[tex]\frac{dv}{dt} = -Ct\\\\dv = -Ctdt\\\\\int\limits^v_{v_0} \, dv = -\int\limits^t_{t_0} \, Ct \\\\v-v_0= -C[\frac{t^2}{2} ]^t_0\\\\v-v_0 = - \frac{1}{2} Ct^2\\\\0 - 20 = - \frac{1}{2}C(8)^2\\\\-20 = -32 C\\\\C = \frac{20}{32} = 0.625 \ m/s^3[/tex]
The acceleration of the object during 8 seconds is calculated as follows;
a = -Ct
a = -0.625(8)
a = -5 m/s²
The distance traveled by the object before stopping in 8 seconds is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\0 = 20^2 + 2(-5)s\\\\0 = 400 - 10s\\\\10s = 400 \\\\s = \frac{400}{10} \\\\s = 40 \ m[/tex]
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A value of C that stops the object in 8 seconds is 1.25 m/s³. The object travels a total distance of 53.33 meters during this time.
An object is moving along the x-axis with an initial velocity of v₀x = 20.0 m/s and an acceleration of ax = -Ct where C is in m/s³. We'll solve for the value of C and the distance traveled in 8.00 s.
Part (a): Finding the value of C
To determine the value of C, consider the velocity function:
v(t) = v₀x + ∫ax dt = v₀x + ∫-Ct dt
Integrating the acceleration to get the velocity:
v(t) = 20.0 m/s - (C/2)t²
Given that the object stops at t = 8.00 s, set v(8.00) = 0:
0 = 20.0 m/s - (C/2)(8.00 s)²
Solving for C:
20.0 m/s = C × 32 s²
C = 40/32 = 1.25 m/s³
Part (b): Distance traveled in 8.00 s
The displacement function x(t) can be found by integrating the velocity function:
x(t) = ∫v(t) dt
x(t) = ∫[20.0 m/s - (C/2)t²] dt
x(t) = 20.0t - (C/6)t³
Using C = 1.25 m/s³ and t = 8.00 s:
x(8.00) = 20.0(8.00) - (1.25/6)(8.00)³
x(8.00) = 160.0 - (1.25/6)(512)
x(8.00) = 160.0 - 106.67
x(8.00) = 53.33 m
In a World Cup soccer match, Juan is running due north toward the goal with a speed of 8.00 m/s relative to the ground. A teammate passes the ball to him. The ball has a speed of 12.0 m/s and is moving in a direction 37.0o east of north, relative to the ground. What are the magnitude and direction of the ball’s velocity relative to Juan?
Answer:
19m/s
22.3 degrees
Explanation:
it is a case aof relative velocity.
the basic for relative velocity vector equation is :
V_b = V_j + V_(b/j)---------------1
V_b: ball velocity relative to ground
V_j : Jaun velocity relative to ground
V_(b/j): ball velocity relative to jaun
reference frame:
We take east and north as +ve x and +ve y
V_(b/j) = V_b - V_j
so for x-axis;
net x-component of V_(b/j) = 12 sin (37) + 0 = 7.22m/s
net y-component of V_(b/j) = 12 cos (37) + 8 = 17.6m/s
magnitude = ((7.22^2)+(17.6^2))^(0.5) = 19 m/s
*direction with respect to Jaun = angle between the vertical (North) and vector V_(b/j)
angle = arctan(7.22/17.6) = 22.3 degrees
A skydiver jumps out of an airplane. Her speed steadily increases until she deploys her parachute, at which point her speed quickly decreases. She subsequently falls to earth at a constant rate, stopping when she lands on the ground.
The question is incomplete but an analysis of the situation using various useful physics concepts can still be made
Answer:
When she immediately jumps out of the plane, the downward force(weight) is greater than any opposing forces upwards (such as air resistance). So the netforce is downwards and therefore the direction of acceleration is also downwards. The direction of acceleration is always in the direction of the netforce The person is not falling at the rate of free fall (9.8 m/s²) because that is for bodies falling in a vacuum and this person is not, air resistance is very much a factor hereUpon deployment of the parachute, upward forces (air resistance) increases matching the downward forces in size, causing the netforce to be zero. A zero netfroce means zero acceleration which is why the person stops accelerating and falls at a constant rateA piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s, what is the magnitude of the centripetal acceleration of the piece of clay on the wheel?
Answer:
[tex]a=24.025\ m/s^2[/tex]
Explanation:
Given that
Distance from the center ,r= 0.1 m
The angular speed ,ω = 15.5 rad/s
We know that centripetal acceleration is given as
a=ω² r
a=Acceleration
r=Radius
ω=angular speed
a=ω² r
Now by putting the values in the above equation we get
[tex]a=15.5^2\times 0.1\ m/s^2[/tex]
[tex]a=24.025\ m/s^2[/tex]
Therefore the acceleration of the clay will be [tex]a=24.025\ m/s^2[/tex].
A woman is riding on a train while watching the display on a GPS unit. She notices that both the "speed" and the "heading" readings are not changing. What can she conclude about the net force acting on the train car?
Answer:
There is no force acting on the body.
Explanation:
While riding a train when a person watches on a GPS that the speed is not changing and also 'heading' is not changing that means the train is moving with constant speed.Heading in GPS means the direction shown in the compass, when the heading is not changing then that means there is no change in the direction of the train.Therefore the person is in a train moving with constant velocity. When any object moves with constant velocity this implies that there is no force acting on the body. This is in accordance with the Newton's first law of motion which states that any particle continues to be in the state of rest or uniform motion until an external force acts upon it.Acceleration is defined in the Newton's second law as the rate of change in velocity. Here we have a constant velocity in terms of both the magnitude and the direction.A turntable reaches an angular speed of "45 rev/min" in "4.10 s" after being turned on. What is its angular acceleration?
Answer:
1.15 rad/s²
Explanation:
given,
angular speed of turntable = 45 rpm
=[tex]45\times \dfrac{2\pi}{60}[/tex]
=[tex]4.71\ rad/s[/tex]
time, t = 4.10 s
initial angular speed = 0 rad/s
angular acceleration.
[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]
[tex]\alpha = \dfrac{4.71-0}{4.10}[/tex]
[tex]\alpha = 1.15\ rad/s^2[/tex]
Hence, the angular acceleration of the turntable is 1.15 rad/s²
A raw egg can be dropped from a third-fl oor window and land on a foam-rubber pad on the ground without breaking. If a 75.0-g egg is dropped from a window located 32.0 m above the ground and a foam-rubber pad that is 15.0 cm thick stops the egg in 9.20 ms, (a) by how much is the pad compressed?(b) What is the average force exertedon the egg after it strikes the pad?
Answer:
N 204.13
Explanation:
Using equation of motion
v² = u² + 2as
u = 0 is the egg was dropped from rest.
v = 2 × 9.8 × 32 = √627.2 = 25.04 m/s
when the egg hit the foam-rubber, the acceleration can be calculated with
a = change in velocity / change in time = - 25.04 / 0.0092 = -2721.74 m/s²
a) how much it is compressed
v² = u² + 2as
- u² = 2 (-2717.4) s
- 627.2 / -5443.48 = s
s = 0.1152 m = 11.52 cm
b) average force exerted on the egg = mΔv / Δt = 25.04 × 0.075 / 0.0092 = 204.13
A sample of N2O gas has a density of 2.697 g/L at 298 K. What must be the pressure of the gas (in mmHg)?
Answer:
[tex]P=1139.16384mmHg[/tex]
Explanation:
Given data
[tex]R=0.08206(\frac{L.Atm}{mol.K} )\\Density=2.697g/L\\Temperature=298K\\f.wt=44(g/mol)\\[/tex]
To find
Pressure
Solution
From Ideal gas law we know that
[tex]PV=nRT\\P=(nR\frac{T}{V} )=(R(\frac{mass}{f.wt} )(\frac{T}{V} ))\\P=R(\frac{mass}{volume}) (\frac{T}{f.wt} )=R(Density)(\frac{T}{f.wt} )[/tex]
Substitute the given values to find pressure
So
[tex]P=(0.08206\frac{L.Atm}{mol.K} )(2.697g/L)(298K)(44g/mol)^{-1}\\ P=1.4989Atm\\[/tex]
Convert Atm to mmHg
Multiply the pressure values by 760
So
[tex]P=1139.16384mmHg[/tex]
You throw a baseball straight up in the air so that it rises to a maximum height much greater than your height. Is the magnitude of the ball’s acceleration greater while it is being thrown or after it leaves your hand? Explain.
The ball's acceleration is constant in magnitude and direction, from the instant it leaves your hand, until the instant it hits the ground, no matter what direction or speed you throw it.
It's the acceleration of gravity, on whatever planet you happen to be standing when you throw the ball.
Final answer:
The baseball's magnitude of acceleration is greater while being thrown than after it leaves the hand due to the additional force applied by the thrower, while in free fall, the ball is subject only to gravity. With air resistance, the ball takes longer to go up than to come back down.
Explanation:
The question pertains to the acceleration of a baseball when thrown straight up into the air. While being thrown, the ball experiences an acceleration greater than the acceleration due to gravity because of the force applied by the person's arm. After the ball leaves the hand, however, the only force acting on it is the force of gravity, which gives it a constant acceleration of approximately 9.81 m/s² downward, regardless of air resistance. In the absence of other forces, the magnitude of acceleration when the ball is in free fall is less than the acceleration imparted to the ball by the thrower's arm.
When air resistance is considered, it acts to slow down the ball as it rises and speeds up as it falls. Therefore, with air resistance, the time it takes for the ball to go up is greater than the time it takes to come back down, because air resistance removes kinetic energy from the ball on the way up, slowing it down more quickly than gravity alone would.
Estimate the number of atoms in your body.
An average adult of weigh 70 kg has approximately [tex]7.0 \times 10^{27}[/tex].
Explanation:
Estimation of Atoms in Human Body
We generally evaluate the amount of elements, liquid, or solid bones in a human body but, we can relate these counts to the atomic level as well. Talking about the approximate amount of Hydrogen, Oxygen and Carbon that we carry in our body, they all make 99% of it neglecting 1% of trace elements.
On the basis of this calculation, we can say that a human body of weight 70 kg has around [tex]7.0 \times 10^{27}[/tex] atoms that are differentiated as 2/3 of hydrogen atoms, 1/4 of Oxygen and only 1/3 of Carbon atoms in the total count.
Calculate the electric field (in V/m) between two parallel plates connected to a 45 V battery if the plates are separated by a) 3 cm and b) 5cm. What is the resulting force on an electron in each of these fields?
Answer:
Explanation:
Given
Voltage of battery [tex]V=45\ V[/tex]
electric field(E) is given by
[tex]E=|-\frac{dV}{dx}|[/tex]
i.e. Change in Potential over a distance x
(a)When Plate are separated by 3 cm apart
[tex]E=|\frac{45}{3\times 10^{-2}}|[/tex]
[tex]E=1500\ V/m[/tex]
(b)When Plates are 5 cm apart
[tex]E=|-\frac{45}{5\times 10^{-2}}|[/tex]
[tex]E=900\ V/m[/tex]
Force on electron is given by
[tex]F=charge\times Electric\ Field[/tex]
[tex]F=q\times E[/tex]
For case a
[tex]F=1.6\times 10^{-19}\times 1500[/tex]
[tex]F=24\times 10^{-17}\ N[/tex]
(b)[tex]F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N[/tex]
A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose mass is about 1/9 and radius 1/2 that of Earth, is mostnearly
A) 1/3 T
B) 2/3T
C)T
D) 3/2T
E) 3 T
Answer:
C)T
Explanation:
The period of a mass-spring system is:
[tex]T=2\pi\sqrt\frac{m}{k}[/tex]
As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.
a 10.0 kg block on a smooth horizontal surface is acted upon by two forces: a horizontal force of 30 N acting to the right and a horizontal force of 60N to the left. What acceleration will the block experience?
Answer:
[tex]a=-3\ m/s^2[/tex]
Explanation:
Second Newton's Law
It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by
[tex]F_n=m.a[/tex]
The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is
[tex]F_n=30\ N-60\ N=-30\ N[/tex]
Thus, the acceleration can be computed by
[tex]\displaystyle a=\frac{F_n}{m}=\frac{-30}{10}=-3\ m/s^2[/tex]
[tex]\boxed{\displaystyle a=-3\ m/s^2}[/tex]
The negative sign indicates the block is accelerated to the left
The block will experience an acceleration of 3.0 m/s² to the left, as calculated by using the net force (30 N leftward) divided by the mass of the block (10.0 kg).
Explanation:To determine the acceleration of a 10.0 kg block on a smooth horizontal surface that is being acted upon by two forces, one must first find the net force acting on the block. In this scenario, there is a 30 N force acting to the right and a 60 N force acting to the left. The net force can be found by subtracting the smaller force from the larger one, which gives us a result of 60 N - 30 N = 30 N acting to the left.
Now, using Newton's second law of motion which states that F = ma (where F is the net force, m is the mass of the object, and a is the acceleration), we can solve for acceleration by rearranging the equation to a = F/m. Plugging the net force (F = 30 N) and the mass of the block (m = 10.0 kg) into the equation yields:
a = 30 N / 10.0 kg = 3.0 m/s²
The acceleration of the block will be 3.0 m/s², and since the net force is to the left, the acceleration will also be directed to the left.
An object is dropped from rest at a height of 128 m. Find the distance it falls during its final second in the air.
Answer:
In the last second, the object traveled 45.6 m.
Explanation:
Hi there!
The equation of the height of the object at a time t is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h =height of the object at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s²).
First, let's find how much time it takes the object to reach the ground. For that, we have to find the value of "t" for which h = 0.
h = h0 + v0 · t + 1/2 · g · t²
Since the object is dropped and not thrown (v0 = 0).
h = h0 + 1/2 · g · t²
0 m = 128 m - 1/2 · 9.8 m/s² · t²
-128 m / -4.9 m/s² = t²
t = 5.1 s
Now, let's find the height of the object 1 s before reaching the ground (at t = 4.1 s):
h = h0 + v0 · t + 1/2 · g · t²
h = 128 m - 1/2 · 9.8 m/s² · (4.1 s)²
h = 45.6 m
Then, in the last second (from 4.1 s to 5.1 s) the object traveled 45.6 m
A submarine periscope uses two totally reflecting 45-45-90 prisms with total internal reflection on the sides adjacent to the 45 degree angles. Explain why the periscope will no longer work if it springs a leak and the bottom prism is covered with water. Note: The index of refraction for water is 1.33. The index of refraction for glass is 1.52
Answer
Given,
Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.
refractive index of water, n_a = 1.33
refractive index of glass, n_g = 1.52
When the light enters the water, water will act as a lens and when we see the object from the periscope the object shown is farther than the usual distance.
A stiff wire bent into a semicircle of radius a is rotated with a frequency f in a uniform magnetic field, as suggested in Fig. 34-51. What are (a) the frequency and (b) the amplitude of the emf induced in the loop
The frequency of the loop is 58 Hz and the amplitude of the emf induced in the loop is 7.73 mV and this can be determined by using the given data.
Given :
A stiff wire bent into a semicircle of radius 'a' is rotated with a frequency f in a uniform magnetic field.
According to the data, the angular speed is 58 rev/sec that is, 364.4 rad/sec, the magnetic field is 15 mT that is, 15 [tex]\times[/tex] [tex]10^{-3}[/tex] T, and the radius 'a' is 3 cm that is, 3 [tex]\times[/tex] [tex]10^{-2}[/tex] m.
a) The frequency is 58 rev/sec that is 58 Hz.
b) The amplitude of the emf induced in the loop can be calculated as:
[tex]\rm \epsilon = \dfrac{\omega B \pi a^2}{2}[/tex]
Now, substitute the values of the known terms in the above formula.
[tex]\epsilon = \dfrac{364.4\times 15\times10^{-3}\times \pi \times (3\times 10^{-2})^2}{2}[/tex]
Further, simplify the above expression.
[tex]\rm \epsilon = 0.007728\;V[/tex]
[tex]\rm \epsilon = 7.73\;mV[/tex]
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A bimetallic strip consists of two 1-mm thick pieces of brass and steel bonded together. At 20C, the strip is completely straight and has a length of 10 cm. At 30C, the strip bends into a circular arc. What is the radius of curvature of this arc?
Answer:
R = 16.67 m
Explanation:
Given:
- Initial Temperature T_i = 20 C
- Thickness of both strips t = 0.001 m
- Final Temperature T_f = 30 C
- Length of the strip L = 0.1 m
- coefficient of linear expansion for brass a_b = 19*10^-6
- coefficient of linear expansion for steel a_s = 13*10^-6
Find:
What is the radius of curvature of this arc R?
Solution:
- The radius of curvature R in relation to dT and a_b, a_s:
R = t / dT*(a_b - a_s)
R = 0.001 / (30-20)*(19-13)*10^-6
R = 16.67 m
A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A package is launched upward, from a point on a roof 10 m above the ground. The initial velocity of the package is 50.5 m/s. Consider all quantities as positive in the upward direction. Does Jim Bond have a chance to catch the package? (calculate how high will it go)
Answer:
The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.
Explanation:
Hi there!
The equation of height and velocity of the package are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the package at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).
v = velocity of the package at a time t.
First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:
v = v0 + g · t
0 = 50.5 m/s - 9.8 m/s² · t
Solving for t:
-50.5 m/s / -9.81 m/s² = t
t = 5.15 s
Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:
h = h0 + v0 · t + 1/2 · g · t²
h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²
h = 140 m
The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.
An ice cube at 0.00 ∘C with a mass of 21.5 g is placed into 500.0 g of water, initially at 31.0 ∘C, in an insulated container. Part A Assuming that no heat is lost to the surroundings, what is the temperature of the entire water sample after all of the ice has melted?
Answer:The temperature of the entire water sample after all of the ice has melted is [tex]30.3^0C[/tex]
Explanation:
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex].......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of ice = 21.5 g
[tex]m_2[/tex] = mass of water = 500.0 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T1 =\text {temperature of ice}= 0^oC[/tex]
[tex]T_2[/tex] = temperature of water =[tex]31.0^oC[/tex]
[tex]c_1[/tex] = specific heat of ice= [tex]2.1J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]21.5\times 2.1\times (T_{final}-0)=-[500.0\times 4.184\times (T_{final}-31.0)][/tex]
[tex]T_{final}=30.3^0C[/tex]
Thus the temperature of the entire water sample after all of the ice has melted is [tex]30.3^0C[/tex]
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 63.0 mm away, making a 2.8 ∘∘ angle with the ground. Ignore all possible aerodynamic effects on the motion of the arrow. Use 9.80 m/s22 for the acceleration due to gravity.
Answer:
[tex]v_0 = 3.53~{\rm m/s}[/tex]
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
[tex]x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}[/tex]
For the y-direction gives
[tex]v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t[/tex]
Combining both equation yields the y_component of the final velocity
[tex]v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}[/tex]
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.
[tex]\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}[/tex]
Final answer:
The question is about projectile motion in Physics, requiring knowledge of equations of motion, range calculation, and trigonometry to find the release angle for an arrow to hit a target and resolve if it will clear an obstacle.
Explanation:
The subject of this question is Physics, specifically dealing with concepts related to projectile motion and the laws of motion. The grade level would likely be High School where students are typically introduced to these topics in a physics curriculum.
As an example, for the question where an archer shoots an arrow at a 75.0 m distant target with an initial speed of 35.0 m/s, one would need to use the equations of projectile motion to solve for the correct release angle. This involves breaking down the initial velocity into its horizontal and vertical components, using the range equation for projectile motion, and applying trigonometry. To determine whether the arrow will go over or under a branch 3.50 m above the release point, one would need to calculate the maximum height of the arrow's trajectory.
In scenarios such as these, understanding the principles of inertia, conservation of momentum, and the conversion of potential energy into kinetic energy are essential.
A uniformly charged, straight filament 6 m in length has a total positive charge of 3 µC. An uncharged cardboard cylinder 1 cm in length and 5 cm in radius surrounds the filament using the filament as its axis of symmetry, with the filament as the central axis of the cylinder. Find the total electric flux through the cylinder. The permittivity of free space is 8.8542 × 10−12 C 2 /N · m2 . 1. 501.959 2. 4356.29 3. 1452.1 4. 1355.29 5. 564.704 6. 250.979 7. 4065.87 8. 141.176 9. 3049.4 10. 847.056 Answer in units of N · m2 /C. 009 (part 2 of 2) 10.0 points What is the electric field at the surface of the cylinder? 1. 224688.0 2. 22468.8 3. 503302.0 4. 209709.0 5. 161776.0 6. 462216.0 7. 719003.0 8. 171191.0 9. 114127.0 10. 179751.0 Answer in units of V/m
Answer:
Part A : 5.) = 564.704 N*m²/C
Part B: 10.) = 179751.0 V/m
Explanation:
A) Applying Gauss'Law to the straight filament, using a cylindrical gaussian surface with the filament as the central axis of the surface, assuming that the electric field is normal to the surface (which means that no flux exist through the lids of the cylinder) and is constant at any point of the surface (except the lids where is 0), we can find the electric flux, as follows:
[tex]E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)[/tex]
where:
r is the radius of the cylinder = 0.05 m
l is the length of the cylinder = 0.01 m
Qenc, is the net charge on the filament enclosed by the gaussian surface
ε₀ = 8.8542*10⁻¹² C²/N*m²
In order to find the value of Qenc, we need to find first the linear charge density, as follows:
[tex]\lambda = \frac{Q}{L} =\frac{+3e-6C}{6m} = 5e-7 C/m[/tex]
The net charge enclosed by the gaussian surface will be just the product of the linear change density λ times the length of the gaussian surface:
[tex]Qenc = \lambda * l = 5e-7C/m * 0.01 m = 5e-9 C[/tex]
According Gauss ' Law, the net flux through the gaussian surface must be equal to the charge enclosed by the surface, divided by the permittivity of free space (in vacuum or air), as follows:
[tex]Flux = \frac{Qenc}{\epsilon 0} = \frac{5e-9C}{8.8542e-12 C2/N*m2} = 564.704 (N*m2)/C[/tex]
which is the same as the option 5.
B) Repeating the equation (1) from above:
[tex]E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)[/tex]
we can solve for E, as follows:
[tex]E = \frac{Qenc}{2*\pi*r*l* \epsilon 0} = \frac{5e-9C}{2*\pi*(0.05m)*(0.01m)*8.8542e-12 C2/N*m2} = 179751.0 V/m[/tex]
which is the same as the option 10 of part B.
As an astronaut visiting the Planet X, you're assigned to measure the free-fall acceleration. Getting out your meter stick and stop watch, you time the fall of a heavy ball from several heights. Your data are as follows:Height (m) Fall Time (s)0 01 0.542 0.723 0.914 1.015 1.17a. Analyze these data to determine the free-fall acceleration on Planet X.b. Determine the uncertainty in the free-fall acceleration.
Answer:
[tex]g \approx 7.4 m/s^2[/tex]
Explanation:
Assuming the following data:
Heigth (m): 0 , 1, 2, 3, 4, 5
Time (s): 0.00, 0.54,0.73, 0.91, 1.01, 1.17
We know from kinematics that the height is given by the following expression:
[tex] h_f = h_i + v_i t + \frac{1}{2} g t^2[/tex]
Assuming for this case that the initial velocity is [tex]v_i[/tex] we can find a polynomial with degree 2 in order to have an estimation for the height with the time.
We can use excel for this and we can see the polynomial adjusted for the data given.
As we can see the best polynomial of degree 2 is given by:
[tex] h(x)= -0.0178 +0.066 x+ 3.6801 x^2[/tex]
For our case x = time and we can rewrite the expression like this
[tex] h(t)= -0.0178 +0.066 t+ 3.6801 t^2[/tex]
And if we are interested on the gravity we want on special the last term of this equation, we can set equal the following terms:
[tex] 3.6801 t^2 = \frac{1}{2} g t^2[/tex]
And solving for g we got:
[tex] 2*3.6801 = g= 7.36 \frac{m}{s^2}[/tex]
So then a good approximation for the gravity of the planet rounded to 2 significant figures is 7.4 m/s^2
To determine the free-fall acceleration on Planet X, you can use the formula: acceleration = 2 * height / fall time^2. Calculating the acceleration for each data point, and the uncertainty can be determined by finding the range of values.
Explanation:a.
To determine the free-fall acceleration on Planet X, we can use the formula: acceleration = 2 * height / fall time^2. We can calculate the acceleration using the given data:
For height 1m, fall time 0.54s, the acceleration is 7.485 m/s^2
For height 2m, fall time 0.72s, the acceleration is 8.660 m/s^2
For height 3m, fall time 0.91s, the acceleration is 9.337 m/s^2
For height 4m, fall time 1.01s, the acceleration is 9.704 m/s^2
For height 5m, fall time 1.17s, the acceleration is 9.335 m/s^2
b.
To determine the uncertainty in the free-fall acceleration, we can calculate the range of values by subtracting the smallest acceleration from the largest acceleration. The smallest acceleration is 7.485 m/s^2 and the largest acceleration is 9.704 m/s^2. Therefore, the uncertainty in the free-fall acceleration is 2.219 m/s^2.
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Calculate the number of atoms contained in a cylinder (1 m radiusand1 m deep)of (a) magnesium (b) lead.
Answer:
The question is incomplete,below is the complete question
"Calculate the number of atoms contained in a cylinder (1μm radius and 1μm deep)of (a) magnesium (b) lead."
Answer:
a. 1.35*10^{11} atoms
b. 1.03*10^{11} atoms
Explanation:
First, we determine the volume of the magnesium in the cylinder container
using the volume of a cylinder
[tex]V=\pi r^{2}h\\ r=10^{-6}m\\ h=10^{-6}m\\V=\pi *10^{-6*2}*10^{-6}\\V=\pi *10^{-18}\\V=3.14*10^{-18}m^{3}\\[/tex]
a. Next we determine the mass of the magnesium ,
using the density=mass/volume
since density of a magnesium
[tex]the density of magnesium =1.738*10^{3}kg/m^{3} \\mass=density * volume \\mass=1.738*10^{3}*3.14*10^{-18}\\mass=5.46*10^{-15}kg\\ \\mass=5.46*10^{-12}g\\[/tex]
Finally to calculate the number of atoms,
we determine the number of moles
mole=mass/molarmass
[tex]mole=5.46*10^{-12}/ 24.305\\mole=0.225*10^{-12}mol\\[/tex]
Hence the number of atoms is
number of atoms=mole*Avogadro's constant
[tex]number of atoms = 0.225*10^{-12}*6.02*10^{23}\\number of atoms =1.35*10^{11} atoms[/tex]
b. for he lead, we determine the mass of the lead ,
using the density=mass/volume
since density of a magnesium
[tex]the density of lead =11.34*10^{3}kg/m^{3} \\mass=density * volume \\mass=11.34*10^{3}*3.14*10^{-18}\\mass=35.60*10^{-15}kg\\ \\mass=35.60*10^{-12}g\\[/tex]
Finally to calculate the number of atoms,
we determine the number of moles
mole=mass/molarmass
[tex]mole=35.60*10^{-12}/ 207.2\\mole=0.1718*10^{-12}mol\\[/tex]
Hence the number of atoms is
number of atoms=mole*Avogadro's constant
[tex]number of atoms = 0.1718*10^{-12}*6.02*10^{23}\\number of atoms =1.03*10^{11} atoms[/tex]
A man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28.0 ∘ above the horizontal. You can ignore air resistance.
The question involves analyzing the projectile motion of a rock thrown from a building using physics principles like motion decomposition and energy conservation. It requires breaking down the initial velocity into horizontal and vertical components and applying kinematic equations to determine parameters such as max height, range, and flight time.
Explanation:This question involves the principles of projectile motion and energy conservation in physics. When the man throws the rock at an angle of 28 degrees above the horizontal with an initial velocity of 26.0 m/s from a height of 14.0 meters, we need to analyze the horizontal and vertical components of the motion separately to determine various aspects of the rock's trajectory, such as its range, maximum height, and time of flight. However, since the specific request is missing in this context, we'll focus on the general approach to solving such problems.
To solve problems involving objects thrown at an angle, we first decompose the initial velocity into its horizontal (vx = v*cos(θ)) and vertical (vy = v*sin(θ)) components, where v is the magnitude of the initial velocity and θ is the angle of projection. The horizontal motion is uniform, meaning the velocity remains constant, whereas the vertical motion is affected by gravity, leading to acceleration in the opposite direction of the initial vertical velocity component.
Energy conservation or kinematic equations can be used to find specific details like maximum height reached, time of flight, and range. For example, the formula s = ut + 0.5at² can be applied where s is displacement, u is initial velocity, a is acceleration (gravity in the case of vertical motion), and t is time. Remember, acceleration due to gravity (a) is -9.8 m/s², indicating it acts downwards. Ignoring air resistance simplifies calculations by omitting drag force considerations.
The maximum horizontal distance is approximately 61.7 meters. This is determined using the projectile motion equations for horizontal distance with initial velocity, angle, and height given.
To find the maximum horizontal distance the rock travels, we can analyze the projectile motion. The initial velocity of 26.0 m/s is broken down into horizontal and vertical components. The horizontal component is [tex]\( v_x = v \cdot \cos(\theta) \)[/tex], where \( v \) is the magnitude of the velocity (26.0 m/s) and \( \theta \) is the angle (28.0 degrees). The vertical component is [tex]\( v_y = v \cdot \sin(\theta) \).[/tex]
The vertical motion is affected by gravity, and the time it takes for the rock to hit the ground can be calculated using the equation [tex]\( h = v_y \cdot t - \frac{1}{2} g t^2 \),[/tex] where \( h \) is the initial height (14.0 m), \( g \) is the acceleration due to gravity (approximately 9.8 m/s\(^2\)), and \( t \) is the time of flight.
Using the quadratic formula to solve for \( t \), we find two solutions: one when the rock is at the initial height and one when it hits the ground. We use the positive solution for the time of flight to calculate the horizontal distance traveled using the equation [tex]\( d = v_x \cdot t \).[/tex]
Substituting the values, we find the maximum horizontal distance to be approximately 61.7 meters.
The question probably maybe: What is the maximum horizontal distance the rock travels before hitting the ground, given that a man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28.0 degrees above the horizontal, ignoring air resistance?
A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the air drag coefficient cD is given by cD= 0.001 m-1. What is the maximum height that the ball reaches? Compare this to the maximum height achieved when air drag is neglected.
Answer:
Explanation:
Given
acceleration is given by
[tex]a=-g-c_Dv^2[/tex]
where [tex]\ddot{y}=a[/tex]
[tex]\dot{y}=v[/tex]
Also acceleration is given by
[tex]a=v\frac{\mathrm{d} v}{\mathrm{d} s}[/tex]
[tex]ds=\frac{v}{a}dv[/tex]
[tex]\int ds=\int \frac{v}{-g-0.001v^2}dv[/tex]
[tex]\Rightarrow Let -g-0.001v^2=t[/tex]
[tex]-0.001\times 2vdv=dt[/tex]
[tex]vdv=-\frac{dt}{0.002}[/tex]
[tex]at\ v_0=50\ m/s,\ t=-g-0.001(50)^2[/tex]
[tex]t=-g-2.5[/tex]
at [tex]v=0,\ t=-g[/tex]
[tex]\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}[/tex]
[tex]\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}[/tex]
[tex]s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}[/tex]
[tex]s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})[/tex]
[tex]s=113.608\ m[/tex]
when air drag is neglected maximum height reached is
[tex]h=\frac{v_0^2}{2g}[/tex]
[tex]h=\frac{50^2}{2\times 9.8}[/tex]
[tex]h=127.55\ m[/tex]
The inner planets formed:
a. by collisions and mergers of planetesimals.
b. in the outer solar system and then were deflected inward by interactions with Jupiter and Saturn.
c. when the Sun's heat destroyed all the smaller bodies in the inner solar system.
d. when a larger planet broke into pieces.
Answer:
a. by collisions and mergers of planetesimals.
Explanation:
Inner planets are planets within 1.5 AU distance from the sun. These are called terrestrial planets because they are somewhat similar to Earth, mainly made of rocks.
The main ingredient of these planets are solar nebula and interstellar dust condensation of which leads to formation of small rock particles. These particles come close to each other under in the influence of gravity and other forces. As the mass of the particles increase they form planetesimals, these planetesimals eventually merge to form planets.
A projectile thrown from a point P moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the projectile could have been thrown. Ignore air resistance.
Answer
70.52°
Explanation
The distance between projectile's position and it's starting point at any time is given by the relation
r² = x² + y²
where x = horizontal distance covered and y = vertical distance covered
According to projectile motion the horizontal displacement is given by
x = v(x)t = v cos(θ) t
Also the vertical component is given by
y = v(y) t - 0.5gt² = v sin(θ) t - 0.5gt²
Substituting the x and y values into the r-equation yields,
r² = (v cos(θ) t)² + (v sin(θ) t - 0.5gt²)²
r² = v²(cos²(θ))t² + v²(sin²(θ))t² – (vg sin(θ))t³+ 0.25 g²(t^4)
r² = v²t² (cos²(θ)+ sin²(θ)) – (vg sin(θ))t³ + 0.25 g²(t^4)
r² = v²t² – (vg sin(θ))t³ + 0.25 g²(t^4)
Differentiate r with respect to t
r(dr/dt) = 2v²t - 3vg sin(θ)t² + g²t³
At maximum angle the projectile could have been thrown above the horizontal, dr/dt = 0
2v²t - 3vg sin(θ)t² + g²t³ = 0
Divide through by t
2v² - 3vg sin(θ)t + g²t² = 0
g²t² - 3vg sin(θ)t + 2v² = 0
This can be solved using the general law for quadratic equations
(-b ± √(b² - 4ac))2a
a = g², b = -3vg sin(θ) c = 2v²
t = ((3vg sin(θ)) ± √(9v²g²sin²(θ) - 8g²v²))/2g²
This equation makes sense when the value under the square root is positive, that is, the square root exists.
9v²g²sin²(θ) - 8g²v² > 0
9sin²(θ) - 8 > 0
Meaning sin²(θ) = 8/9
Sin θ = (2√2)/3
θ = 70.52°
QED!!!
"The velocity of a diver just before hitting the water is -10.0 m/s, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 1.16 s of the dive?
Answer:
Explanation:
Given
velocity of diver [tex]u=-10\ m/s[/tex] i.e. downward motion
acceleration due to gravity [tex]a=g=-9.8\ m/s^2[/tex]
time [tex]t=1.16\ s[/tex]
using equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
[tex]y=(-10)\cdot 1.16-\frac{1}{2}(-9.8)(1.16)^2[/tex]
[tex]y=-11.6-6.593[/tex]
[tex]y=-18.19\ m[/tex]
I.e. in downward direction
The displacement of the diver during the last 1.16 seconds of her dive is -11.6 meters. The negative sign indicates a downward movement.
Explanation:In physics, displacement is the overall change in position of an object. It is calculated by multiplying velocity and time. In this case, the velocity of the diver is -10.0 m/s (a negative sign indicating downward motion) and the time is 1.16 s. To find the displacement, multiply the velocity and the time: (-10.0 m/s) × (1.16 s) = -11.6 m. The minus sign still indicates downward direction, and it means that the diver moved 11.6 meters downward in the last 1.16 seconds of her dive.
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