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OLga [1]
3 years ago
15

A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin

g velocity, until a docking maneuver is completed and the vehicle is attached to the station. How does the total momentum of the docked vehicle and station compare to the momentum of each object before the docking maneuver? greater momentum same momentum less momentum
Physics
1 answer:
NeX [460]3 years ago
3 0

Answer:

c. same momentum

Explanation:

got it correct on edge2020

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Some of the energy the car gained as it was pulled was lost because of heat from friction. The rest of the energy was transforme
Lisa [10]

Answer:

454,320 joules

Explanation:

The work done on an object is equal to its change in kinetic energy: Change in KE = F × d.

Plug the values for F and d into the formula and solve:

Change in KE = 2,524 × 180

= 454,320 joules

The roller coaster gains 454,320 joules of energy from the work done on it by the chain.

7 0
3 years ago
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While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the
olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0
3 years ago
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
Hey guys how to prevent from shock
Pie
U need an insulator to prevent the shock passing through , that's why there is a plastic sheet covering the copper wires so that it can prevent the electricity passing through as plastic is an insulator and does not conduct electricity
8 0
3 years ago
A radar operates at a wavelength of 3 centimeters. what is the frequency of the these waves? the speed of light is 3 × 108 m/s .
aliya0001 [1]
We are given information:
λ = wavelength = 3cm = 0.03m \\ 
c = speed of light =  3* 10^{8}m/s

Formula that connects wavelength and frequency is:
frequency= \frac{speed of light}{wavelength} \\ f= \frac{3* 10^{8}}{0.03} \\ f=1 * 10^{10} Hz = 10GHz
4 0
3 years ago
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