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ycow [4]
3 years ago
5

Practice questions, will mark brainliest!

Physics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

266.67Watts

Explanation:

Time = 2.5hr to seconds

3600s = 1hr

2.5hrs = 3600×2.5= 9000s

Force = 32N

Distance = 75km  to m

1000m = 1km

75km = 1000×75 = 75000m

Power = workdone / time

Work = force × distance

Therefore work = 32N × 75000m

Work = 2400000Nm

Power = work ➗ time

Power = 2400000Nm ➗ 9000s

Power = 266.67Watts

Watts is the S. i unit of power

I hope this was helpful, please mark as brainliest

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If the total mass hanging on one end of the string that creates the tension in the string is 0.300 kg , what is the mass density
soldi70 [24.7K]

Answer:

The mass density of the string is (0.3/L)kg/m

Explanation:

Mass density of the string = Mass/Length

Mass = 0.3kg

The length of the string is unknown so it is assumed to be L meter(s)

Therefore, mass deny of the string = 0.3kg/Lm = (0.3/L)kg/m

7 0
4 years ago
Which statement best defines work? (2 points) Group of answer choices
Irina18 [472]

Answer:last choice is correct. Work is equally to force times distance

Explanation:to generalize, W is equal to to force as a function of distance integrated over distance. Work is equivilent to energy. See work/force equivalency

4 0
2 years ago
Explain how Newton's third law is applied when a rocket ship is being launched?
garri49 [273]

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6 0
3 years ago
A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what
Fudgin [204]

Answer:

v = 40 [m/s].

Explanation:

Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.

P=m*v\\

where:

m = mass [kg]

v = velocity [m/s]

P =20*10\\P =200 [kg*m/s]

Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.

v = P/m\\v = 200/5\\v = 40 [m/s]

3 0
3 years ago
A jet airplane is in level flight. The mass of the airplane is m=9010kg. The airplane travels at a constant speed around a circu
Lyrx [107]

Answer:

The magnitude of the lift force L = 92.12 kN

The required angle is ≅ 16.35°

Explanation:

From the given information:

mass of the airplane = 9010 kg

radius of the airplane R = 9.77 mi

period T = 0.129 hours = (0.129 × 3600) secs

= 464.4 secs

The angular speed can be determined by using the expression:

ω = 2π / T

ω = 2 π/ 464.4

ω = 0.01353 rad/sec

The direction \theta = tan^{-1} ( \dfrac{\omega ^2 R}{g})

\theta = tan^{-1} ( \dfrac{0.01353 ^2 \times (9.77\times 1609)}{9.81})

θ = 16.35°

The magnitude of the lift force  L = mg ÷ Cos(θ)

L = (9010 × 9.81) ÷ Cos(16.35)

L = 88388.1  ÷ 0.9596

L = 92109.32 N

L = 92.12 kN

3 0
3 years ago
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