Destinations humans have traveled in space

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

Law Incorporation [45]

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:

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

3 0

3 years ago

Help<br> A. 12,240<br> B. 6,120

nata0808 [166]

Answer:

A. 12,240.

Explanation:

1,530 times 8.0 = 12,240.

hope this helped :)

3 0

2 years ago

A student walks 10 m North, 6 m South, 7 m West and 4 m East. Find the student's distance AND displacement

BlackZzzverrR [31]

Answer: Distance: 27m Displacement: 7m

Explanation: Distance is total, Displacement is from the start.

Hope this helped!

Mark Brainliest if you feel like it!

5 0

3 years ago

What is the coefficient of friction between the skates and the ice?

natta225 [31]

The coefficient of friction is 0.051

Explanation:

The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2 - u^2 = 2as$

where:

v = 0 is the final velocity of the skater (he comes to a stop)

u = 10.0 m/s is his initial velocity

a is the acceleration

$s=1.0\cdot 10^2 m = 100 m$ is the distance he travels before stopping

Solving for a, we find the acceleration of the skater:

$a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2$

We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):

$F= ma = -\mu mg$

where

$-\mu mg$ is the force of friction

m is the mass of the skater

$\mu$ is the coefficient of friction

$a=-0.5 m/s^2$ is the acceleration

$g=9.8 m/s^2$ is the acceleration of gravity

Solving for $\mu$ , we find the coefficient of friction:

$\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

8 0

3 years ago

Which statement best describes the electrons in this illustration ?

Nonamiya [84]

A. electrons are negatively charged outside nucleus

7 0

3 years ago

Read 2 more answers