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lord [1]
2 years ago
10

A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

xperiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed Vesc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density p 3.84 x108 g/m3 and volume V 2.17 x 1012 m3 Recall that the universal gravitational constant is G 6.67 x 10-11 N m2/kg2
Physics
1 answer:
Blizzard [7]2 years ago
6 0
<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity V_{esc} is given by the following equation:

V_{esc}=\sqrt{\frac{2GM}{R}}   (1)

Where:

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M  is the mass of the asteroid

R  is the radius of the asteroid

On the other hand, we know the density of the asteroid is \rho=3.84(10)^{8}g/m^{3} and its volume is V=2.17(10)^{12}m^{3}.

The density of a body is given by:

\rho=\frac{M}{V}  (2)

Finding M:

M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})  (3)

M=8.33(10)^{20}g=8.33(10)^{17}kg  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

V=\frac{4}{3}\piR^{3}   (5)

Finding R:

R=\sqrt[3]{\frac{3V}{4\pi}}   (6)

R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}   (7)

R=8031.38m   (8)  This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}   (9)

Finally:

V_{esc}=117.626m/s This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

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As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
f = (\frac{g + v_{r}}{g + v_{s}})f_{o} -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
v_{r} = Velocity of the receiver/observer relative to the medium = ?.
v_{s} = Velocity of the source with respect to medium = 0 m/s.
f_{o} =  Frequency emitted from source = 400 Hz.
f = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

408 = ( \frac{340 + v_{r}}{340 + 0})*400

\frac{408}{400} =  \frac{340 + v_{r}}{340}

Solving above would give you,
v_{r} = 6.8 m/s

The correct answer = 6.8m/s



7 0
2 years ago
The potential energy of a 40 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
Fynjy0 [20]
The formula for potential energy is PE=mgh
It can have that high of a potential energy if it's relative height what super high.
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2 years ago
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
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2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg
harkovskaia [24]

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given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 ×10^{-2} m

separation L = 0.5 mm = 0.5 ×10^{-3} m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 ×10^{-2}  / 11

θ = 0.234

and we know diffraction grating for max

L sinθ  = mλ

here we know m = 1  so put all value and find λ

L sinθ  = mλ

0.5 ×10^{-3}  sin(0.234)  = 1 λ

λ = 2042.02 ×10^{-9}  m

λ = 2042 nm

3 0
2 years ago
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