Ask question

Ask question

All categories

2 years ago

13

A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

at would keep the cable taut. A motorized climber could slowly carry rockets to the top, where they could be launched away from the earth using much less energy.
What would be the escape speed for a craft launched from a space elevator at a height of 56,000 km?

1 answer:

Valentin [98]2 years ago

0 0

Answer:

The escape speed for the craft is 1.49 m/s.

Explanation:

In this case we need to find the escape speed for a craft launched from a space elevator at a height of 56,000 km. The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{d}}$

Here,

G is universal gravitational constant

M is mass of earth

d = r + h, r is radius of Earth

$v=\sqrt{\dfrac{2\times 6.67\times 10^{-11}\times 5.972 \times 10^{24}}{(6,371 \times 10^3\times 56000\times 10^3)}}\\\\v=1.49\ m/s$

So, the escape speed for the craft is 1.49 m/s.

You might be interested in

What is the instantaneous velocity of a freely falling object 9.0 s after it is released from a position of rest? Express your a

Umnica [9.8K]

Answer:

So instantaneous velocity after 9 sec will be 88.2 m/sec

Explanation:

We have given time t = 9 sec

As the object is released from rest so its initial velocity u = 0 m/sec

We have to find its final velocity v

Acceleration due to gravity $g=9.8m/sec^2$

From first equation of motion we know that $v=u+gt$

$v=0+9.8\times 9=88.2m/sec$

So instantaneous velocity after 9 sec will be 88.2 m/sec

5 0

2 years ago

A woman makes 50% more than her husband. Together they make $2500 each month? How much does the woman earn each month?a) $950b)

Sergeeva-Olga [200]

Answer:b

Explanation:

Given

Woman earn 50% more than her husband

Total sum of their money is $=\$ 2500$

Suppose man earns $\$ P$

so women earns $P+0.5 P=P(1+0.5)=1.5 P$

Sum of their money is

$P+1.5 P=2500$

$P=\$ 1000$

Women earns $=1.5\times 1000=\$ 1500$

6 0

1 year ago

A boat is headed with a velocity of 18 meters/second toward the west with respect to the water in a river. If the river is flowi

IrinaK [193]

If the boat's velocity is 18m/sec relative to the water in the river and not the shore, it would need to be added the river speed of 2.5m/sec to get a total of 20.5m/sec. The 20.5m/sec would then be the total velocity of the boat relative to the shore. From personal experience, I know that when one runs with the tide, one is adding the tide flow speed to one's boat speed (what it would be in neutral waters) to get a sometimes much faster speed.

7 0

2 years ago

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is $v = 350 \ m/s$

Explanation:

From the question we are told that

The distance of separation is d = 4.00 m

The distance of the listener to the center between the speakers is I = 5.00 m

The change in the distance of the speaker is by $k = 60 cm = 0.6 \ m$

The frequency of both speakers is $f = 700 \ Hz$

Generally the distance of the listener to the first speaker is mathematically represented as

$L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}$

$L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}$

$L_1 = 5.39 \ m$

Generally the distance of the listener to second speaker at its new position is

$L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}$

$L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}$

$L_2 = 5.64 \ m$

Generally the path difference between the speakers is mathematically represented as

$pD = L_2 - L_1 = \frac{n * \lambda}{2}$

Here $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v}{f}$

=> $L_2 - L_1 = \frac{n * \frac{v}{f}}{2}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves

=> $5.64 - 5.39 = \frac{1 * v}{2*700}$

=> $v = 350 \ m/s$

7 0

2 years ago

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

2 years ago