the viscosity of water at 0°C is 2.12 × [tex]10^−1[/tex] poise (P).
To express the viscosity of water at 0°C (2.12×[tex]10^−2[/tex]kg/(s⋅m)) in poise (P), we need to convert the units accordingly.
We know that 1 P = 1 g/(s⋅cm), and we have the viscosity in kg/(s⋅m). To convert kg to g and m to cm, we need to multiply the value by appropriate conversion factors.
1 kg = 1000 g (since there are 1000 grams in a kilogram)
1 m = 100 cm (since there are 100 centimeters in a meter)
Using these conversion factors, we can convert the viscosity:
2.12 × [tex]10^−2[/tex] kg/(s⋅m) = 2.12 ×[tex]10^−2[/tex] × 1000 g/(s⋅100 cm)
= 2.12 ×[tex]10^−2[/tex] × 10 g/(s⋅cm)
= 2.12 ×[tex]10^−1[/tex] g/(s⋅cm)
Therefore, To express the viscosity of water at 0°C (2.12×[tex]10^−2[/tex] kg/(s⋅m)) in poise (P), we need to convert the units accordingly.
We know that 1 P = 1 g/(s⋅cm), and we have the viscosity in kg/(s⋅m). To convert kg to g and m to cm, we need to multiply the value by appropriate conversion factors.
1 kg = 1000 g (since there are 1000 grams in a kilogram)
1 m = 100 cm (since there are 100 centimeters in a meter)
Using these conversion factors, we can convert the viscosity:
2.12 ×[tex]10^−2[/tex]kg/(s⋅m) = 2.12 ×[tex]10^−2[/tex] × 1000 g/(s⋅100 cm)
= 2.12 × [tex]10^−2[/tex] × 10 g/(s⋅cm)
= 2.12 × 10^−2[tex]10^−2[/tex] g/(s⋅cm)
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A cube is suspended by a string from the ceiling such that half of its volume is submerged in water. What can you say about the string's tension when it is half submerged compared to the weight of the cube ? [The density of water is rho=1000 kg/m3 and the density of the cube is rho=2000 kg/m3.] The tension force is exactly half the weight of the cube. The tension force is more than half the weight of the cube, but less than the full weight: The tension force is less than half the weight of the cube, but more than 1/3 rd the weight of the cube. The tension force is exactly 1/3 ird the weight of the cube The tension force is less than 1/3 rd the weight of the cube
The tension force in the string when the cube is half submerged is less than half the weight of the cube, but more than 1/3rd the weight of the cube.
When an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by the object.
In this case, since half of the cube's volume is submerged, it displaces an amount of water with a weight equal to half of its own weight.
The tension in the string must counterbalance the downward force of the cube's weight and the upward buoyant force. Since the buoyant force is equal to half the weight of the cube, the tension force in the string must be less than that to maintain equilibrium. However, it must be more than 1/3rd the weight of the cube since the cube is still partially submerged.
Therefore, the tension force in the string is less than half the weight of the cube but more than 1/3rd the weight of the cube.
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A hypothetical spherical planet of radius R and its density varies as rho=Kr, where K is constant and r is the distance from the center. Determine the pressure caused by gravitational pull inside( r
The density inside the spherical planet of radius R and density varying as rho=Kr is given byrho=Kr (r < R)Since the planet is spherical, we can use Gauss' law to calculate the gravitational field inside the planet.
The Gaussian surface is a sphere of radius r with its center at the center of the planet. By symmetry, the gravitational field is constant on this surface, and its magnitude is given by$$g=\frac{GM_r}{r^2},$$where Mr is the mass enclosed within the sphere of radius r. Therefore, the gravitational force acting on an element of area dA on the Gaussian surface is$$dF=gdA.$$The mass enclosed within the Gaussian surface is given by$$M_r=\int_0^r 4\pi r^2 \rho dr=\int_0^r 4\pi K r^3 dr=\frac{4}{3} \pi K r^4.$$Substituting this expression into the expression for the gravitational field gives$$g=\frac{GM_r}{r^2}=\frac{4}{3} \pi G K r^2.$$Therefore,
the pressure caused by gravitational pull inside the planet is given by$$P=-\frac{dU}{dV}=\frac{d}{dV} \left( -\frac{3}{2} \frac{GM_r^2}{5R^5} \right) = \frac{3}{2} \frac{GM_r^2}{5R^6}.$$Substituting the expression for Mr, we get$$P=\frac{3}{2} \frac{G^2 K^2 r^8}{5R^6}.$$
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Inside a radiation therapy device, an electron is in the presence of a uniform electric field with a magnitude of 305 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s2)
(b) The electron is initially at rest. What is its speed (in m/s) after 7.50 ✕ 10−9 s?
The magnitude of the acceleration of the electron is 3.34797e+32 m/s².
The speed of the electron after 7.50 ✕ 10−9 s is 2.24097e+31 m/s.
(a) The magnitude of the acceleration of the electron is:
a = E / m = 305 N/C / 9.11e-31 kg
a = 3.34797e+32 m/s²
where:
a is the acceleration of the electron (m/s²)
E is the magnitude of the electric field (N/C)
m is the mass of the electron (kg)
(b) The electron's speed after 7.50 ✕ 10−9 s is:
v = at = 3.34797e+32 m/s² * 7.50e-9 s
v = 2.24097e+31 m/s
where:
v is the speed of the electron (m/s)
a is the acceleration of the electron (m/s²)
t is the time (s)
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Find the total resistance (in Ohms) for the circuit in the figure given that: R1 = 1Ω, R2 = 6 Ω and R3 = 12 Ω.
Your answer should be a number with one decimal place, do not include the unit.
The total resistance for the circuit, when R1 = 1Ω, R2 = 6Ω, and R3 = 12Ω, is approximately 4.4Ω.
To find the total resistance for the circuit, we need to determine the equivalent resistance when the resistors R1, R2, and R3 are combined.
In the given circuit, R1 and R2 are connected in series, and the resulting equivalent resistance (Rs) is the sum of their individual resistances:
Rs = R1 + R2 = 1Ω + 6Ω = 7Ω
The resistor R3 is connected in parallel to the combination of R1 and R2. To calculate the equivalent resistance (Rp) of the parallel combination, we use the formula:
1/Rp = 1/R3 + 1/Rs
Substituting the values, we have:
1/Rp = 1/12Ω + 1/7Ω
Simplifying the expression:
1/Rp = (7 + 12)/(12 * 7) = 19/84
To find Rp, we take the reciprocal of both sides:
Rp = 84/19 ≈ 4.421Ω
Therefore, the total resistance for the circuit, when R1 = 1Ω, R2 = 6Ω, and R3 = 12Ω, is approximately 4.4Ω.
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A cannon ball is launched at an angle of 27∘ relative to horizontal. What is the vertical component of the projectile's velocity, if the cannon ball leaves the cannon at 80 m/s ?
The vertical component of the projectile's velocity is approximately 36.3 m/s. To find the vertical component of the projectile's velocity, given that a cannonball is launched at an angle of 27∘ relative to the horizontal and leaves the cannon at 80 m/s, we will use trigonometric functions.
We can determine the vertical component of the velocity (v) using the formula:v = v₀sin(θ) where v₀ is the initial velocity and θ is the angle of elevation. Here, v₀ = 80 m/s and θ = 27∘.
Substituting the values into the equation above, we have:v = 80 m/s sin(27∘)
Using a calculator, we find that sin(27∘) ≈ 0.454
Therefore, the vertical component of the projectile's velocity is:v ≈ 80 m/s × 0.454 ≈ 36.3 m/s
Therefore, the vertical component of the projectile's velocity is approximately 36.3 m/s.
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A car traveling at 33 m/s runs out of gas while traveling up a 9.0
∘
slope. Part A How far will it coast before starting to roll back down? Express your answer in meters.
The car will coast 264.61 m before starting to roll back down.
The given information is;
The initial velocity of the car is 33 m/s.
The car runs out of gas while traveling up a 9.0° slope.
a) We need to find how far the car will coast before starting to roll back down.
To solve this problem, first, we will find the distance traveled by the car before coming to rest.
So,
The distance traveled by the car before coming to rest can be calculated as;
v² = u² + 2as
0 = 33²/2 × g × sin9°
0 = 1089 / (2 × 9.8 × 0.15643)
= 1089 / 3.062
= 355.48 m
Now we can calculate the distance it will cover while coasting upwards. The car's velocity will be zero at the highest point of the slope.
So, the potential energy at the highest point will be converted into kinetic energy when the car starts to roll back down.
Distance covered = h
= u² / 2g
= (355.48 sin9°)² / (2 × 9.8)
= 264.61 m
Thus, the car will coast 264.61 m before starting to roll back down.
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A box with mass m sits stationary on the floor. You push on the box with a
force of magnitude F at an angle of θ below the horizontal. The coefficient of static friction
between the box and the floor is µs.
(a) What is the minimum value of F to get the box to start sliding? Your answer should
be in terms of m, θ, and µs.
(b) There is a maximum angle θmax such that if the angle is any larger, no magnitude of
force is large enough to get the box to start sliding. Determine θmax in terms of m and µs
To get a stationary box to start sliding on the floor, the minimum value of the applied force F needs to be µs * m * g * cos(θ), where µs is the coefficient of static friction, m is the mass of the box, g is the acceleration due to gravity, and θ is the angle below the horizontal at which the force is applied.
(a)The minimum value of F required to get the box to start sliding is **F = µs * m * g * cos(θ)**, where g is the acceleration due to gravity.
To overcome static friction and initiate sliding, the applied force F must be equal to or greater than the maximum static friction force. The maximum static friction force is given by:
Maximum static friction force = µs * Normal force
The normal force acting on the box is equal to the weight of the box, which is m * g, where m is the mass and g is the acceleration due to gravity. The vertical component of the applied force is F * sin(θ), and it balances the weight of the box. Therefore:
m * g = F * sin(θ)
Solving this equation for F:
F = (m * g) / sin(θ)
However, since we need the minimum value of F to start sliding, the force F must overcome both the vertical and horizontal components of the static friction force. The horizontal component is µs * Normal force * cos(θ) = µs * m * g * cos(θ). Therefore, the minimum value of F is µs * m * g * cos(θ).
(b) The maximum angle θmax, such that no magnitude of force is large enough to get the box to start sliding, can be determined by equating the horizontal component of the applied force to the maximum static friction force.
µs * Normal force * cos(θmax) = m * g * sin(θmax)
Dividing both sides of the equation by cos(θmax):
µs * Normal force = m * g * tan(θmax)
The maximum value of the coefficient of static friction is 1, so:
µs * Normal force = m * g * tan(θmax) ≤ m * g
µs * m * g ≤ m * g
Simplifying the equation:
µs ≤ 1
Therefore, the maximum angle θmax is such that the coefficient of static friction (µs) is less than or equal to 1.
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You have a parallel-plate 5.59×10
−6
F capacitor that is charged to 0.00803C. While the capacitor is isolated, you change the plate separation so that the capacitance becomes 1.47×10
−6
F. How much work W do you perform in this process?
The work performed in changing the plate separation of the capacitor is approximately 2.75 Joules. For, the work performed in changing the plate separation of the capacitor, we can use the formula
W = (1/2) × (C2 × V2^2 - C1 × V1^2)
where W represents the work done, C1 and C2 represent the initial and final capacitance respectively, V1 and V2 represent the initial and final voltage respectively.
C1 = 5.59 × 10^(-6) F,
C2 = 1.47 × 10^(-6) F,
Q = 0.00803 C (charge on the capacitor).
We know that the charge Q is equal to the product of capacitance and voltage:
Q = C × V.
Rearranging the equation, we can solve for V:
V = Q / C.
Now, we can calculate the initial and final voltages:
V1 = Q / C1,
V2 = Q / C2.
Substituting these values into the work formula, we have:
W = (1/2) × (C2 × (Q / C2)^2 - C1 × (Q / C1)^2).
Simplifying the equation:
W = (1/2) × (Q^2 / C2 - Q^2 / C1).
Factoring out Q^2:
W = (1/2) × Q^2 × (1 / C2 - 1 / C1).
Substituting the given values:
W = (1/2) × (0.00803 C)^2 × (1 / (1.47 × 10^(-6) F) - 1 / (5.59 × 10^(-6) F)).
Calculating the value:
W ≈ 2.75 J.
Therefore, the work performed in changing the plate separation of the capacitor is approximately 2.75 Joules.
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Required information Margie, who weighs \( 531 \mathrm{~N} \), is standing on a bathroom scale that weighs \( 48.0 \mathrm{~N} \). With what magnitude force does the scale push up on Margie?
A bathroom scale measures the force between it and whatever is on it. This force is the same as the magnitude of the gravitational force between the scale and the object. Let's denote the force by F, the mass of the object by m, and the acceleration due to gravity by g.
Then we have:F = mgThe force on the scale is simply the normal force exerted by the scale on the object. If the object is on a flat, horizontal surface, then the normal force is just equal and opposite to the gravitational force:
N = F = mgHere, Margie weighs \(531\ N\) and the bathroom scale weighs [tex]\(48.0\ N\)[/tex].
Since the force is a vector, it has magnitude and direction. Since the scale pushes up, we know that the direction of the force is up. Hence, the scale pushes up on Margie with a magnitude equal to the normal force, which is equal and opposite to the gravitational force.
Therefore, the magnitude of the force that the scale pushes up on Margie is:N = F = mg= (531 + 48.0) N= 579.0 NThus, the bathroom scale pushes up on Margie with a force of 579.0 N.
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On a spacecraft, two engines are turned on for 937 s at a moment when the velocity of the craft has x and y components of v
0x
=5580 m/s and v
0y
=6440 m/s. While the engines are firing, the craft undergoes a displacement that has components of x=3.79×10
6
m and y =5.80×10
6
m. Find the (a) x and (b) y components of the craft's acceleration. (a) Number Units (b) Number Units
The x-component of acceleration is approximately -1535.36 m/s², and the y-component of acceleration is approximately -251.89 m/s².
To find the x and y components of the spacecraft's acceleration, we can use the kinematic equation:
Δv = aΔt
where Δv is the change in velocity, a is the acceleration, and Δt is the time interval.
Given:
Initial x-component velocity (v0x) = 5580 m/s
Initial y-component velocity (v0y) = 6440 m/s
Displacement in x-direction (Δx) = 3.79 * 10⁶ m
Displacement in y-direction (Δy) = 5.80 * 10⁶ m
Time interval (Δt) = 937 s
(a) To find the x-component of the acceleration (ax):
Δvx = axΔt
Δvx = vxf - v0x
We can calculate vxf (final x-component velocity) using the displacement in the x-direction:
vxf = Δx / Δt
Substituting the given values:
vxf = (3.79 * 10⁶ m) / (937 s)
vxf ≈ 4044.64 m/s
Now, we can calculate the x-component of the acceleration:
Δvx = vxf - v0x
Δvx = 4044.64 m/s - 5580 m/s
Δvx ≈ -1535.36 m/s
Therefore, the x-component of the spacecraft's acceleration is approximately -1535.36 m/s².
(b) To find the y-component of the acceleration (ay):
Δvy = ayΔt
Δvy = vyf - v0y
We can calculate vyf (final y-component velocity) using the displacement in the y-direction:
vyf = Δy / Δt
Substituting the given values:
vyf = (5.80 * 10⁶ m) / (937 s)
vyf ≈ 6188.11 m/s
Now, we can calculate the y-component of the acceleration:
Δvy = vyf - v0y
Δvy = 6188.11 m/s - 6440 m/s
Δvy ≈ -251.89 m/s
Therefore, the y-component of the spacecraft's acceleration is approximately -251.89 m/s².
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The actual question is:
On a spacecraft, two engines are turned on for 937 s at a moment when the velocity of the craft has x and y components of v₀x =5580 m/s and v₀y =6440 m/s. While the engines are firing, the craft undergoes a displacement that has components of x=3.79×10⁶ m and y =5.80×10⁶ m.
a) Find the x components of the craft's acceleration.
b) Find the y components of the craft's acceleration.
The form of a sound wave traveling through air is s(x,t)=s
m
cos(kx+3πt), where x is in meters and t in seconds. What is the shortest time interval any air molecule takes along the path to move between displacements s=+ 0.86 s
m
and s=−0.86 s
m
? a. 0.22 s b. 0.42 s c. 0.30 s d. 0.53 s e. 0.27 s
The shortest time interval any air molecule takes along the path to move between displacements of +0.86 m and -0.86 m in the given sound wave is 0 seconds.
In the given wave equation s(x,t) = sm * cos(kx + 3πt), the argument of the cosine function, (kx + 3πt), needs to change by 2π radians for the air molecule to move between the specified displacements. However, after solving the equation, it is found that the difference in time, t2 - t1, is zero. Therefore, the air molecule takes no time to move between these displacements. None of the provided options (a, b, c, d, e) is the correct answer.
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A toy rocket, launched from the ground, rises vertically with an acceleration of 22 m/s2 for 10 s until its motor stops.
Disregarding any air resistance, what max- imum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s2 .
The maximum height attained by the rocket is 610 m above the ground.
Acceleration of the rocket = 22 m/s²
Time taken = 10 seconds
Acceleration due to gravity = 9.8 m/s²
To find the maximum height achieved by the rocket, we use the formula for displacement:
[tex]�=��+12��2s=ut+ 21 at 2[/tex]
Where:
s = maximum height attained
u = initial velocity = 0 (since the rocket was launched from rest)
a = acceleration = 22 - 9.8 = 12.2 m/s² (since the acceleration of gravity acts in the opposite direction to the rocket's acceleration)
t = time taken = 10s
Putting the values in the equation, we get:
[tex]�=0×10+12×12.2×102s=0×10+ 21 ×12.2×10 2 �[/tex]
[tex]=0+610s=0+610�=610�s=610m[/tex]
Therefore, the maximum height attained by the rocket is 610 m above the ground.
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A projectile is shot from the ground directly upwards and passes from a point that is at a height of 95 m from the ground. The time interval between the two passes from this point is 18 seconds. Find the initial velocity of the projectile and the maximum height of its trajectory. Use g = 10 m/s2.
The initial velocity of the projectile is 30 m/s, and the maximum height of its trajectory is 145 m.
Let's denote the initial velocity of the projectile as "v" (in m/s). When the projectile reaches its maximum height, its final velocity becomes zero. Using the kinematic equation, we can calculate the time taken for the projectile to reach its maximum height.
The first pass from the point at a height of 95 m occurs when the projectile is moving upward. The time taken to reach this point can be determined using the equation: s = ut + (1/2)[tex]at{^2[/tex], where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get 95 = vt - (1/2)[tex]gt^2[/tex].
The second pass occurs when the projectile is falling downward. The time taken to reach the same point can be calculated using the same equation, considering the negative acceleration due to gravity. This time would be 18 seconds more than the first pass, so we have 95 = 0 - (1/2)g[tex](t + 18)^2[/tex].
Solving these two equations simultaneously, we can find the initial velocity of the projectile, which is v = 30 m/s. Substituting this value into the equation for the first pass, we can find the time taken to reach the maximum height, which is approximately 3 seconds.
Using the equation v = u + gt, we find the final velocity at the maximum height is -30 m/s. Again using the equation [tex]v^2[/tex] = [tex]u^2[/tex] + 2as and solving for s, we can find the maximum height to be 145 m.
Therefore, the initial velocity of the projectile is 30 m/s, and the maximum height of its trajectory is 145 m.
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We shoot an object from the ground with initial velocity v0 that forms an angle θ with the horizontal. Find this angle θ if it is known that the range is 4 times the maximum height of the trajectory.
To find the angle θ, we can utilize the range formula and the maximum height formula of projectile motion. The range formula is given by:
R = ([tex]v0^2[/tex]* sin 2θ) / g ----- (1)
And the maximum height formula is given by:
H = ([tex]v0^2[/tex]* [tex]sin^2[/tex]θ) / (2g) ----- (2)
Given that the range is 4 times the maximum height, we can express this relationship in equation form as:
R = 4H
Substituting the expression for H in terms of v0 and θ from equation (2) into the above equation, we obtain:
([tex]v0^2[/tex] * sin 2θ) / g = 4 * [([tex]v0^2[/tex]* [tex]sin^2[/tex]θ) / (2g)]
Simplifying this equation, we have:
sin 2θ = 8 [tex]sin^2[/tex]θ
Further simplifying:
2 sin θ cos θ = 8 [tex]sin^2[/tex]θ
Dividing both sides by sin θ, we get:
cot θ = 4
Taking the inverse cotangent ([tex]cot^-1[/tex]) on both sides, we can determine the value of θ:
θ = [tex]cot^-1[/tex](4)
Calculating the inverse cotangent of 4, we find:
θ ≈ 14.04 degrees
Therefore, the angle θ is approximately 14.04 degrees.
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10.0 kg pendulum bob is placed on a 5.00 m long string and pulled back 6.00
∘
. What is the period of the pendulum when it is released? 12.3 s 8.80 s 3.215 4.49 s What is the intensity of a sound wave if the power output of the source is 3.56 W and the surface area 1.58 m
2
? 1.98 W/m
2
0.444 W/m
2
2.25 W/m
2
5.62 W/m
2
What is the relative intensity of a sound wave with an intensity of 0.000845 W/m
2
? 89.3 dB 8.93 dB −151 dB −15.1 dB
The period of the pendulum with a 10.0 kg bob on a 5.00 m string and a 6.00° displacement is approximately 8.80 seconds (b). The intensity of the sound wave with a power output of 3.56 W and a surface area of 1.58 m² is approximately 2.25 W/m² (c). The relative intensity of the sound wave with an intensity of 0.000845 W/m² is approximately 8.93 dB (b).
To calculate the period of the pendulum, we use the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum string, and g is the acceleration due to gravity. By applying the given values, including the effective length of the pendulum calculated using L_eff = L(1 - cosθ), we find the period to be approximately 8.80 seconds (b).
The intensity of a sound wave is determined by the formula I = P/A, where I is the intensity, P is the power output, and A is the surface area. By substituting the given values, the intensity is calculated to be approximately 2.25 W/m² (c).
The relative intensity of a sound wave is measured in decibels (dB) and can be found using the formula β = 10log(I/I₀), where β is the relative intensity, I is the given intensity, and I₀ is the reference intensity. By plugging in the provided intensity value, we find the relative intensity to be approximately 8.93 dB (b).
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A train has a length of 108 m and starts from rest with a constant acceleration at time t=0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time t=13.6 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t=39.1 s, the car is again at the rear of the train. Find the magnitudes of (a) the car's velocity and (b) the train's acceleration.
We are given the following data: A train has a length of 108 m and starts from rest with a constant acceleration at time t=0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time t=13.6 s, the car just reaches the front of the train.
Ultimately, however, the train pulls ahead of the car, and at time t=39.1 s, the car is again at the rear of the train. To solve this problem, we can assume that the train starts moving with a constant acceleration and at a time t=0 s, the train is at rest.
The distance between the car and the train is 108 m. At time t = 13.6 s, the car reaches the front of the train. So, the distance covered by the car in 13.6 seconds is the same as the length of the train: 108 m = (u × 13.6) + (0.5 × a × 13.6²)
The above equation gives us the value of u + 96
a. Similarly, when the car reaches the back of the train at time t = 39.1 s, the distance travelled by the car is the same as the length of the train:108 m = (u × 39.1) + (0.5 × a × 39.1²)
The above equation gives us the value of u + 760.55a. Now we have two equations and two unknowns. Therefore, we can solve the equations simultaneously to obtain the values of u and a.
108 m = (u × 13.6) + (0.5 × a × 13.6²)108 m = (u × 39.1) + (0.5 × a × 39.1²)
Solving the above equations simultaneously, we get: u = 28.01 m/sa = 0.114 m/s²
Now that we have found the value of acceleration, we can calculate the car's velocity at any given time. Since the car is moving with a constant velocity, its velocity will remain the same. Therefore, the car's velocity is: v = 28.01 m/s
The train's acceleration is: a = 0.114 m/s²
The problem is related to the motion of a train and a car moving on a straight path. We are given the length of the train, and the car is moving with a constant velocity. At time t = 0, the train starts moving with a constant acceleration. We need to find the magnitudes of the car's velocity and the train's acceleration. To solve this problem, we can use the equations of motion, which relate the distance travelled by an object to its initial velocity, acceleration, and time. We know that the train starts from rest, so its initial velocity is 0.
The car, on the other hand, is moving with a constant velocity, which means its acceleration is 0. Using the equations of motion, we can relate the distance covered by the car to the distance covered by the train. At time t = 13.6 s, the car reaches the front of the train. Therefore, the distance covered by the car in 13.6 seconds is the same as the length of the train. Similarly, when the car reaches the back of the train at time t = 39.1 s, the distance travelled by the car is again the same as the length of the train. We can use these two equations to solve for the car's velocity and the train's acceleration.
We found that the car's velocity is 28.01 m/s, and the train's acceleration is 0.114 m/s². This means that the train is accelerating at a very slow rate, and it takes a long time for the train to catch up to the car. Ultimately, however, the train pulls ahead of the car, indicating that its acceleration is greater than the car's velocity.
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(a) What is the area and uncertainty in area of one side of a rectangular metal slab that has a length of (21.4±0.4) cm and a width of (9.8±0.1)cm ? (Give your answers in cm
2
.) (b) What If? If the thickness of the slab is (1.1±0.2)cm, what is the volume of the slab and the uncertainty in this volume? (Give your answers in cm
3
.) (बल) x± बन्र x)cm
3
Based on the data provided, (a) area of one side of the rectangular metal slab is (210.92 ± 1.39) cm² and (b) the volume of the slab is (232.01 ± 4.24) cm³.
(a) Area of rectangular metal slab is given by the formula : A = lw
where l is the length and w is the width of the rectangular metal slab.
Substituting l = (21.4 ± 0.4) cm and w = (9.8 ± 0.1) cm in the above formula, we get :
A = (21.4 ± 0.4) cm x (9.8 ± 0.1) cm
On expanding the above expression, we get :
A = (21.4 x 9.8) ± [(0.4/21.4)² + (0.1/9.8)²]¹/²
A = (210.92 ± 1.39) cm²
Therefore, area of one side of the rectangular metal slab is (210.92 ± 1.39) cm².
(b) The volume of the slab is given by the formula : V = Al
where A is the area and l is the thickness of the rectangular metal slab.
Substituting A = (210.92 ± 1.39) cm² and l = (1.1 ± 0.2) cm in the above formula, we get :
V = (210.92 ± 1.39) cm² x (1.1 ± 0.2) cm
On expanding the above expression, we get : V = 232.01 ± 4.24 cm³
Therefore, the volume of the slab is (232.01 ± 4.24) cm³.
Thus, the correct answers are : (a) (210.92 ± 1.39) cm² ; (b) (232.01 ± 4.24) cm³
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question 7 options: a photon with 9.01x10-19 joules of energy has a wavelength of nanometers.
The photon with an energy of [tex]9.01 * 10^{-19}[/tex] J has a wavelength of approximately 220.3 nm.
To find the wavelength of a photon given its energy, we can use the equation:
E = hc/λ
Where:
E is the energy of the photon,
h is the Planck's constant (approximately 6.62607015 × 10^-34 J·s),
c is the speed of light in a vacuum (approximately 2.998 × 10^8 m/s),
λ is the wavelength of the photon.
Rearranging the equation to solve for λ, we have:
λ = hc/E
Given:
[tex]E = 9.01 * 10^{-19}[/tex] J
Substituting the known values:
[tex]\lambda = \frac{(6.62607015 * 10^{-34} * 2.998 * 10^8 )}{(9.01 * 10^{-19}}[/tex]
[tex]\lambda = \frac{(1.98644591 * 10^{-25})}{(9.01 * 10^{-19})}[/tex]
[tex]\lambda \approx 2.203 * 10^{-7} m[/tex]
To convert this wavelength to nanometers, we can multiply by 10⁹:
=λ ≈ 220.3 nm.
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The ink drops have a mass m = 1.00×10−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 24.0 m/s . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length D0 = 2.30 cm , where there is a uniform vertical electric field with magnitude E = 7.80×104 N/C . If a drop is to be deflected a distance d = 0.340 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m3 , and ignore the effects of gravity.
1. The electric force (F) can be calculated using the formula: F = q * E . 2. Determine the distance the ink drop is deflected (d) by the time it reaches the end of the deflection plate. 3. Equate the electric force to the force causing the deflection. 4. The magnitude of charge q must be zero for the ink drop to be deflected a distance d = 0.340 mm by the time it reaches the end of the deflection plate.
To find the magnitude of charge q that must be given to the ink drop, we can use the following steps:
1. Calculate the electric force acting on the ink drop as it passes through the uniform vertical electric field between the deflecting plates. The electric force (F) can be calculated using the formula:
F = q * E
Where q is the charge on the ink drop and E is the magnitude of the electric field.
2. Determine the distance the ink drop is deflected (d) by the time it reaches the end of the deflection plate.
3. Equate the electric force to the force causing the deflection. In this case, the force causing the deflection is the horizontal component of the gravitational force (mg). Since we are ignoring the effects of gravity, this force will be zero. Therefore, we have:
F = 0
q * E = 0
4. Solve the equation for the magnitude of charge q:
q = 0 / E
q = 0
This means that the magnitude of charge q must be zero for the ink drop to be deflected a distance d = 0.340 mm by the time it reaches the end of the deflection plate.
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A waterball is dropped from the top of a 42.3 m balcony in order to hit a walking person below. As the person is leaning over to watch the balloon fall, their glasses fall off 2.88 s after the waterballoon was dropped.
How high above the ground are the glasses when the water balloon hit the unsuspecting walking person?
Assume the acceleration for both the waterball and the glasses is 9.81m/s2, the acceleration due to gravity.
The glasses were about 94.1m above the ground when the water balloon hit the unsuspecting walking person.Distance travelled by waterball before glasses fall off=42.3m.
Time taken by glasses to fall=2.88s, Acceleration due to gravity=9.81m/s²
Formula used:S = ut + (1/2)at² where S is the distance, u is initial velocity, t is the time taken, and a is acceleration.
Let's find out how high above the ground are the glasses when the water balloon hits the unsuspecting walking person.
Distance travelled by waterball before glasses fall off = 42.3m, Time taken by glasses to fall = 2.88s, Acceleration due to gravity = 9.81m/s² Acceleration due to gravity is negative (-9.81m/s²) because it is in the opposite direction to the movement of the object (upwards).
For the glasses to fall, their initial velocity is zero.So, the formula used for the glasses is:S = (1/2)at²S = (1/2) × 9.81m/s² × (2.88s)²S = 39.7m.
Thus, the glasses were 39.7m above the ground when the water balloon hit the unsuspecting walking person.
Distance travelled by waterball before glasses fall off = Distance travelled by waterball after glasses fall off.
The initial velocity of the water balloon is zero when it is dropped from the balcony.
So, the formula used to calculate the distance travelled by the waterball is:S = (1/2)at²S = (1/2) × 9.81m/s² × (t)²S = 4.905t².
Also, Distance travelled by the waterball before glasses fall off + Distance travelled by the waterball after glasses fall off = Total distance travelled by waterball = 42.3m.
So, the distance travelled by the waterball after the glasses fall off is:42.3m - Distance travelled by waterball before glasses fall off=42.3m - 4.905t² .
On equating the above two equations, we get:4.905t² = 42.3m - 4.905t²9.81t² = 42.3mt² = 4.314s.
The total time taken by the water balloon to reach the ground is 4.314s.
Therefore, the height of the glasses, when the water balloon hits the unsuspecting walking person, is:S = (1/2)at²S = (1/2) × 9.81m/s² × (4.314s)²S = 94.1m (approx.).
Thus, the glasses were about 94.1m above the ground when the water balloon hit the unsuspecting walking person.
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the sum
A
+
B
=
C
, vector
A
has a magnitude of 11.6 m and is angled 38.0
∘
cocinterdockwise from the +x direction, and vector
C
has a magnitude of 15.7 m and is angled 21.3 " counterclockwise from the - x direction. What are (a) the magnitude and (b) the angle (relative to +x) of
B
? State your angle as a positive number. (a) Number Units (b) Number Units
The magnitude of vector B is approximately 7.3 m, and its angle relative to the +x direction is approximately 96.7° counterclockwise.
In this problem, we are given vector A and vector C, and we need to find vector B. Vector A has a magnitude of 11.6 m and is angled 38.0° counterclockwise from the +x direction. Vector C has a magnitude of 15.7 m and is angled 21.3° counterclockwise from the -x direction.
To find vector B, we first need to determine its magnitude. We can use the given information that the sum of vectors A and B is equal to vector C. Using the Pythagorean theorem, we can write:
[tex]|A|^2 + |B|^2 = |C|^2[/tex]
Substituting the known values, we have:
[tex](11.6)^2 + |B|^2 = (15.7)^2[/tex]
Simplifying the equation gives:
[tex]135.36 + |B|^2 = 246.49[/tex]
Solving for |B|, we find:
|B| ≈ √(246.49 - 135.36) ≈ 7.3 m
Next, we need to determine the angle of vector B relative to the +x direction. We can use trigonometry to solve for the angle. Since we know the magnitude of vector B, we can use the arctangent function to find the angle:
θ = arctan[tex](B_y / B_x)[/tex]
Substituting the known values, we have:
θ = arctan(7.3 * sin(96.7°) / 7.3 * cos(96.7°))
Calculating the value gives:
θ ≈ arctan(-1.34) ≈ -50.99°
Since the question asks for the angle as a positive number, we convert the angle to its positive equivalent:
θ ≈ 360° - 50.99° ≈ 309.01°
Therefore, the magnitude of vector B is approximately 7.3 m, and its angle relative to the +x direction is approximately 96.7° counterclockwise.
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Consider five different objects for which the mas and the velocity are given. Rank the object in terms of their kinetic energy from high to low. That is, the item with the most kinetic energy should be ranked 1st and the the item with the least amount of kinetic energy should be last.
M=0.1 kg,V=5 m/s
M=0.4 kg,V=1 m/s
M=0.3 kg,V=2 m/s
M=0.2 kg,V=4 m/s
M=0.5 kg,V=5 m/s
The objects ranked in terms of their kinetic energy from high to low are:
1. Object 5 (0.5 kg, 5 m/s)
2. Object 4 (0.2 kg, 4 m/s)
3. Object 3 (0.3 kg, 2 m/s)
4. Object 2 (0.4 kg, 1 m/s)
5. Object 1 (0.1 kg, 5 m/s)
Using the formula KE = 1/2 mv², we can calculate the kinetic energy of each object:
1. KE = 1/2 x 0.5 kg x (5 m/s)² = 6.25 J
2. KE = 1/2 x 0.2 kg x (4 m/s)² = 1.6 J
3. KE = 1/2 x 0.3 kg x (2 m/s)² = 0.6 J
4. KE = 1/2 x 0.4 kg x (1 m/s)² = 0.2 J
5. KE = 1/2 x 0.1 kg x (5 m/s)² = 1.25 J
Therefore, the objects ranked in terms of their kinetic energy from high to low are:
1. Object 5 (0.5 kg, 5 m/s)
2. Object 4 (0.2 kg, 4 m/s)
3. Object 3 (0.3 kg, 2 m/s)
4. Object 2 (0.4 kg, 1 m/s)
5. Object 1 (0.1 kg, 5 m/s)
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A 16kHz voltage is applied to a 30pF capacitor, and 9.3 mA of RMS current is measured. Determine the reactive power? The result should be in VAR Calculate Answers to 2 decimal points
To determine the reactive power, we need to calculate the impedance of the capacitor and then use the formula for reactive power.
Step 1: Calculate the impedance of the capacitor using the formula:
Z = 1 / (2πfC), where Z is the impedance, f is the frequency, and C is the capacitance.
In this case, the frequency is 16kHz (16,000 Hz) and the capacitance is 30pF (30 * 10^-12 F).
Plugging in these values, we get:
Z = 1 / (2 * 3.14 * 16000 * 30 * 10^-12)
Z = 1 / (2 * 3.14 * 480 * 10^-12)
Z = 1 / (3012.96 * 10^-12)
Z = 332.61 Ω (approximately)
Step 2: Calculate the reactive power using the formula:
Q =[tex]I^2 * X[/tex], where Q is the reactive power, I is the RMS current, and X is the reactance.
The reactance (X) is the imaginary part of the impedance, which is given by X = √(Z^2 - R^2), where R is the resistance (assumed to be zero in this case).
Plugging in the values, we get:
X = √(332.61^2 - 0^2)
X = √(110686.9921)
X = 332.61 Ω (approximately)
Now, we can calculate the reactive power:
Q = (9.3 * 10^-3)^2 * 332.61
Q = 0.08679 VAR (approximately)
Therefore, the reactive power is approximately 0.08679 VAR.
In summary, to calculate the reactive power, we first calculate the impedance of the capacitor using the frequency and capacitance values. Then, we find the reactance by taking the square root of the impedance squared minus the resistance squared. Finally, we use the RMS current and reactance to calculate the reactive power using the formula.
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Please select the word from the list that best fits the definition
Calculated by T = Fd sin θ
Torque is the word from the list that best fits the given equation, and it represents the twisting or turning force applied to an object, calculated using the formula T = Fd sin θ.
The word from the list that best fits the definition of the equation T = Fd sin θ is "Torque." In physics, torque is a measure of the turning or twisting force applied to an object. It is calculated by multiplying the magnitude of the force (F) applied to an object by the perpendicular distance (d) from the point of rotation to the line of action of the force, and then multiplying that by the sine of the angle (θ) between the force vector and the lever arm.
The equation T = Fd sin θ represents the mathematical relationship for calculating torque. T represents torque, F represents the force, d represents the distance, and θ represents the angle between the force and the lever arm. Torque is a fundamental concept in mechanics and is used to describe rotational motion and the effectiveness of a force in causing an object to rotate. It is commonly used in fields such as engineering, physics, and mechanics, particularly in the study of machines, levers, and rotational dynamics.
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What conclusions can be made about the image forming properties of (i) positive or concave mirrors; and (ii) negative or convex mirrors.
(i) Positive or concave mirrors:The image formed by positive or concave mirrors depends on the object's location relative to the mirror. The mirror reflects and converges the rays of light.
The image formed can either be real or virtual depending on the object's location. When the object is located beyond the center of curvature, a real inverted image is formed.
The image formed is real, inverted, and reduced in size.The image formed by positive or concave mirrors is larger than the actual object when it is placed between the focal point and the center of curvature. If the object is placed between the focal point and the mirror, a virtual, erect, and magnified image is formed.
(ii) Negative or convex mirrors: The image formed by a convex mirror is always virtual, erect, and smaller in size. The rays of light that pass through a convex mirror diverge. As a result, no real image is formed. A virtual image is formed behind the mirror. When an object is placed in front of a convex mirror, the reflected image is smaller than the object. This is due to the fact that the image formed by a convex mirror is always smaller than the object.
It is also referred to as a virtual image since it cannot be projected on a screen. Thus, the image formed by a convex mirror is always virtual and reduced in size. This is because the mirror diverges the light that passes through it.
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4. A Pelton turbine produces 8 MW of power under net head of 440 m. The turbine has two jets. The buckets deflect the jet by an angle of 165°. If the bucket friction coefficient is 0.85, compute the following:
(a) Discharge
(b) Diameter of each jet
(c) The total force exerted by the jets on the wheel in the tangential direction
(d) Power produced by the runner
(e) Hydraulic efficiency
The overall efficiency, coefficient of velocity and speed ratio may be assumed
0.85, 0.98, and 0.47, respectively.
a) The discharge is [tex]1,847.83 m^3/s[/tex], b) the diameter of each jet is 10.49 m, c) the total force exerted by the jets is 100,536,960 N, d) the power produced by the runner is 5,483,204,556.8 W, e) and the hydraulic efficiency is 67.4%.
(a) The discharge can be calculated using the formula
Q = P / (ρgh),
where Q is the discharge, P is the power produced, ρ is the density of water, g is the acceleration due to gravity, and h is the net head.
Plugging in the given values:
Q = 8,000,000 / (1000 * 9.81 * 440) = [tex]1,847.83 m^3/s[/tex].
(b)The diameter of each jet can be determined using the formula
[tex]d = (4Q / (\pi v))^{0.5},[/tex]
where d is the diameter, Q is the discharge, and v is the velocity of each jet. The velocity can be found using the coefficient of velocity (cv) and the speed ratio (u), given as
[tex]v = cv * u * (2gh)^{0.5}[/tex].
Plugging in the given values:
v = 0.98 * 0.47 * (2 * 9.81 * 440)^0.5 = 54.48 m/s.
Substituting this into the diameter formula:
[tex]d = (4 * 1,847.83 / (\pi * 54.48))^{0.5} = 10.49 m[/tex].
(c) The total force exerted by the jets can be calculated using the formula
F = ρQv,
where F is the force, ρ is the density of water, Q is the discharge, and v is the velocity of each jet.
Plugging in the given values:
F = 1000 * 1,847.83 * 54.48 = 100,536,960 N.
(d)The power produced by the runner can be determined using the formula
P = F * vt,
where P is the power, F is the force, and vt is the tangential velocity. Since the force is exerted tangentially, vt is the same as the velocity of each jet. Therefore,
P = 100,536,960 * 54.48 = 5,483,204,556.8 W.
(e)The hydraulic efficiency (ηh) can be calculated using the formula ηh = P / (ρgQh).
Plugging in the given values:
ηh = 5,483,204,556.8 / (1000 * 9.81 * 1,847.83 * 440) = 0.674 or 67.4%.
In conclusion, the discharge is [tex]1,847.83 m^3/s[/tex], the diameter of each jet is 10.49 m, the total force exerted by the jets is 100,536,960 N, the power produced by the runner is 5,483,204,556.8 W, and the hydraulic efficiency is 67.4%.
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On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 4.60 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 41.3 m/s at an angle of 28
∘
above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball? (a) Number Units (b) Number Units
To solve this problem, we can use the equations of projectile motion.
Let's break down the given information:
On the distant planet:
- The ball is launched with a speed of 41.3 m/s.
- The launch angle is 28 degrees above the horizontal.
- The ball lands at the same level as the tee.
- The ball travels 4.60 times farther than it would on Earth.
We need to find:
(a) The maximum height of the ball.
(b) The range of the ball.
Let's start by calculating the values on Earth and then adjust them for the distant planet.
(a) Maximum height:
On Earth, the maximum height can be calculated using the formula:
H = (v^2 * sin^2θ) / (2 * g)
Where:
H = Maximum height
v = Initial velocity (41.3 m/s)
θ = Launch angle (28 degrees)
g = Acceleration due to gravity on Earth (approximately 9.8 m/s^2)
Plugging in the values, we get:
H = (41.3^2 * sin^2(28)) / (2 * 9.8)
Now, to find the maximum height on the distant planet, we need to multiply this value by 4.60:
H_distant_planet = 4.60 * H
(b) Range:
On Earth, the range can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
Where:
R = Range
Plugging in the values, we get:
R = (41.3^2 * sin(2 * 28)) / 9.8
Now, to find the range on the distant planet, we need to multiply this value by 4.60:
R_distant_planet = 4.60 * R
Let's calculate the values:
(a) Maximum height:
H = (41.3^2 * sin^2(28)) / (2 * 9.8)
H_distant_planet = 4.60 * H
(b) Range:
R = (41.3^2 * sin(2 * 28)) / 9.8
R_distant_planet = 4.60 * R
Now we can compute these values:
(a) Maximum height:
H = (41.3^2 * sin^2(28)) / (2 * 9.8)
H_distant_planet = 4.60 * H
Calculating H on Earth:
H = (41.3^2 * sin^2(28)) / (2 * 9.8)
H ≈ 37.656 m
Now, calculating the maximum height on the distant planet:
H_distant_planet = 4.60 * H
H_distant_planet ≈ 173.0976 m
(b) Range:
R = (41.3^2 * sin(2 * 28)) / 9.8
R ≈ 157.999 m
Now, calculating the range on the distant planet:
R_distant_planet = 4.60 * R
R_distant_planet ≈ 726.394 m
Therefore, on the distant planet:
(a) The maximum height of the ball is approximately 173.1 meters.
(b) The range of the ball is approximately 726.4 meters.
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A standard I kilogram weight is a cylinder 54.0 mm in height and 41.5 mm in diameter. What is the density of the material? kg/m
3
Answer:
The density of the material is approximately 5,549 kg/m³.
The standard 1 kilogram weight is a cylinder with dimensions of 54.0 mm in height and 41.5 mm in diameter.
The first step in determining the density of the material is to find its volume.
Using the formula for the volume of a cylinder,
V = πr²h
where V is the volume,
r is the radius, and
h is the height.
We can find the radius by dividing the diameter by
2.r = 41.5 mm / 2r
= 20.75 mm
Now we can calculate the volume.
V = πr²hV
= π(20.75 mm)²(54.0 mm)V
≈ 1.801 x 10⁵ mm³
We want the density to be in kg/m³, so we need to convert the volume from mm³ to m³.
1 m = 1000 mm1 m³
= (1000 mm)³
= 10⁹ mm³V
= 1.801 x 10⁵ mm³ x (1 m³ / 10⁹ mm³)V
≈ 1.801 x 10⁻⁴ m³
Now that we have the volume in m³, we can calculate the density.
The formula for density is:
density = mass / volume
We know that the mass of the weight is 1 kg, so we can substitute that in.
density = 1 kg / (1.801 x 10⁻⁴ m³)
density ≈ 5,549 kg/m³
Therefore, the density of the material is approximately 5,549 kg/m³.
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You wake up on an unknown planetoid. You have a 2 m long pendulum and note that it swings with a period of 10 seconds. Using sticks and their shadows at different locations around the planet, you determine that the radius of your planet is 5.4
∗
10
5
m. What is the mass of your planet?
The mass of the planet is 2.43 × 10²³ kg.
The formula that is used for this problem is given by: T = 2π√(l/g)
Where, T = Period of oscillation of the pendulum, l = Length of the pendulum, g = Acceleration due to gravity On substituting the given values, we get:
T = 10sl = 2mg/5.4 * 10^5 m
Hence, we getg = 4π²l/T² × (5.4 * 10⁵ m)
On substituting the values of l and T, we get: g = 8.62 m/s².
To calculate the mass of the planet, we use the formula given below: g = (GM)/R²
Where, G = Gravitational constant = 6.67 × 10⁻¹¹ N(m/kg)²
M = Mass of the planet, R = Radius of the planet, On substituting the given values and solving for M, we get:
M = (gR²)/G
M = (8.62 m/s² × (5.4 × 10⁵ m)²)/(6.67 × 10⁻¹¹ N(m/kg)²)
M = 2.43 × 10²³ kg.
Therefore, the mass of the planet is 2.43 × 10²³ kg.
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The plates of parallel plate capacitor A consist of two metal discs of identical Side view radius, R
1
=3.37 cm, separated by a distance d=3.77 mm, as shown in the figure. a) Calculate the capacitance of this parallel plate capacitor with the space between the plates filled with air. Capacitor A Tries 0/100 b) A dielectric in the shape of a thick-walled cylinder of outer radius R
1
=3.37 cm, inner radius R
2
=1.87 cm, thickness d=3.77 mm, and dielectric constant k= 2.97 is placed between the plates, coaxial with the plates, as shown in the figure. 2.97 is placed between the plates, coaxial with the plates, as Calculate the capacitance of capacitor B, with this dielectric. Metal disc Calculate the capacitance of capacitor B, with this dielectric. Capacitor B Tries 0/100 c) The dielectric cylinder is removed, and instead a solid disc of radius R
1
made of Dielectric the same dielectric is placed between the plates to form capacitor C, as shown in the figure. What is the new capacitance? Tries 0/100 Capacitor C Dielectric
Capacitance, C = 2.2 × 10⁻¹¹ F for (a)
Capacitance with the dielectric is 1.5 × 10⁻¹⁰ F for (b)
The new capacitance of capacitor C is 7.4 × 10⁻¹¹ F for (c)
a) Capacitance of parallel plate capacitor A when the space between the plates is filled with air:
Side view radius, R₁ = 3.37 cm
The separation between the plates, d = 3.77 mm = 0.377 cm
The permittivity of free space, ε₀ = 8.85 × 10⁻¹² F/m
Capacitance is given by, C = ε₀A/d
Where A is the area of each plate. Area of each plate, A = πR₁² = π (3.37 × 10⁻²)²
Therefore, capacitance, C = ε₀A/d = (8.85 × 10⁻¹² × π × (3.37 × 10⁻²)²)/ (0.377 × 10⁻²)= 2.2 × 10⁻¹¹ F
b) Capacitance of capacitor B when a dielectric in the shape of a thick-walled cylinder is placed between the plates:
Side view radius, R₁ = 3.37 cm
Inner radius, R₂ = 1.87 cm
Thickness, d = 3.77 mm = 0.377 cm
Dielectric constant, k = 2.97
Let the capacitance of capacitor B be Cᵇ
The area of each plate with the dielectric in place is given by, A = π(R₁² - R₂²)
Capacitance with the dielectric is given by, Cᵇ = kε₀A/d= kε₀π(R₁² - R₂²)/d= 2.97 × 8.85 × 10⁻¹² × π × [(3.37 × 10⁻²)² - (1.87 × 10⁻²)²]/(0.377 × 10⁻²)= 1.5 × 10⁻¹⁰ F
c) Capacitance of capacitor C when a solid disc of radius R₁ made of the same dielectric is placed between the plates:
Let the capacitance of capacitor C be C.C.
The area of each plate with the dielectric disc in place is given by, A = πR₁²
Capacitance with the dielectric disc is given by, C.C = kε₀A/d= kε₀πR₁²/d= 2.97 × 8.85 × 10⁻¹² × π × (3.37 × 10⁻²)²/(0.377 × 10⁻²)= 7.4 × 10⁻¹¹ F
Therefore, the new capacitance of capacitor C is 7.4 × 10⁻¹¹ F.
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