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3 years ago

If the mass of an object is 5 kg and the velocity is 8 m/s, what is the momentum?

Physics

1 answer:

topjm [15]3 years ago

3 0

Answer:

40 kg•m/s

Explanation:

momentum (p) = mass (kg) × velocity (m/s)

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How much GPE is stored when an 80kg astronaut climbs to the top of a 5m high lunar lander? The gravity strength on the moon is 1

8_murik_8 [283]

Answer:

The GPE, stored is 640 Joules

Explanation:

The given parameters are;

The given mass of the astronaut, m = 80 kg

The height of the top of the lunar lander to which the astronaut climbs, h = 5 m

The gravity strength on the moon, g = 1.6 N/kg

The Gravitational Potential Energy, GPE, stored is given according to the following equation;

GPE stored = m·g·h

Therefore, by substituting the known values, we have;

GPE Stored = 80 kg × 1.6 N/kg × 5 m = 640 Joules

The GPE, stored = 640 Joules.

6 0

3 years ago

Conclusion for ohm law

kari74 [83]

Ohm’s Law states that electrical current is proportional to voltage and therefore inversely proportional to resistance - V = I x R

7 0

3 years ago

If a car accelerates uniformly from rest to 15 meters

Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0

3 years ago

Read 2 more answers

n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a

Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0

3 years ago

Read 2 more answers

What is the process of preparing a person to perform strenuous work and to recover from that work as quickly as possible?

vesna_86 [32]

The process of preparing a person to perform strenuous work and to recover from that work as quickly as possible is called training, I bet. Such w<span>arm up exercises as like stretching and other should lead person to the recovery. Hope it will help.</span>

3 0

4 years ago