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Elza [17]
3 years ago
7

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

had a mass of 110 kg, how much would you weigh when standing on the surface of this asteroid? Express your answer to two significant figures and include appropriate units. 2. If you could walk on the surface of this asteroid, what minimum speed would you need to launch yourself into an orbit just above the surface of the asteroid? Express your answer to two significant figures and include appropriate units.
Physics
1 answer:
Law Incorporation [45]3 years ago
3 0

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

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nika2105 [10]
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3 years ago
A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente
inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

F=\frac{mv^2}{r}

Where, m is the mass of the ball, v is the speed and r is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

20.4 = \frac{3.1\times v^2}{2}

and, v^2 = 13.16

Thus, v = 3.63 m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = \frac{\text {Path length} }{\text {Speed}} = \frac{12.56}{3.63} = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0
3 years ago
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
A ball with a momentum of 16 kg.M/s strikes a ball at rest. What is the total momentum of both the balls after the collision.
melomori [17]

Answer:

Total momentum = 16 Kgm/s

Explanation:

Let the momentum of the two balls be A and B respectively.

Momentum A = 16 kgm/s

Momentum B = 0 kgm/s (since the ball is at rest).

Total momentum = A + B

Total momentum = 16 + 0

Total momentum = 16 Kgm/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

8 0
3 years ago
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What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.
Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0
3 years ago
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