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2 years ago

What is the change in momentum of a 50. kg woman goes from running at 8.0 m/s to 0.0 m/s?

Physics

1 answer:

oksian1 [2.3K]2 years ago

4 0

Answer:

400 kg-m/s

Explanation:

Given that,

Mass of the woman, m = 50 kg

Initial velocity, u = 8 m/s

Final velocity, v = 0 m/s

We need to find the change in momentum of the woman. It can be given by :

$\Delta p=m(v-u)\\\\=50\times (8-0)\\\\=400\ kg-m/s$

So, the change in momentum is 400 kg-m/s.

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Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush

Sergio039 [100]

Answer:

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

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2 years ago

If you use a force of 90 N to pick up a 10 pound bag of charcoal, what is the acceleration?

hjlf

Answer:

9ms^2

Explanation:

since ,Force=mass*acceleration

then, acceleration=force/mass

and, Force=90N

Mass=10pound

therefore, acceleration=90/10

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3 years ago

Example of energy transfer from potential to kinetic and back

Alex_Xolod [135]

Swimmer and Divers. The Potential energy is transferred into Kinetic energy, and allows the diver to submerge into the water. The Kinetic energy then allows the diver to submerge and dive into the water. Potential energy however, is needed to allow the diver to get back out of the water after diving to get up and go and dive again, and then the Kinetic energy is transferred back to Potential energy to repeat the process.

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2 years ago

Both mechanical and electromagnetic waves

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Answer:

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Explanation:

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3 years ago

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during

Katarina [22]

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

The speed increases from 0 m/s to 1341 m/s.
The time elpased is 2.0 min.
1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

$a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s} = 11 m/s^{2}$

Learn more: brainly.com/question/16274121

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2 years ago