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steposvetlana [31]
3 years ago
15

The movement of energy from the sun toward the earth is an example of ?

Physics
2 answers:
Anni [7]3 years ago
8 0

Answer:

I took the quiz, Radiation

I am Lyosha [343]3 years ago
3 0

Gravitational  potential energy

is the answer

trust me :)

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a rocket initially at rest on the ground lifts off vertically with a constant acceleration of 2.0 x 10^1 meters per second^2. Ho
xenn [34]

Here's the formula for the distance covered by an accelerating body in some amount of time ' T '. This formula is incredibly simple but incredibly useful. It pops up so often in Physics that you really should memorize it:

D = 1/2 a T²

Distance = (1/2)·(acceleration)·(time²)

This question gives us the acceleration and the distance, and we want to find the time.

(9,000 m) = (1/2) (20 m/s²) (time²)

(9,000 m) = (10 m/s²) (time²)

Divide each side by 10 m/s²:

(9,000 m) / (10 m/s²) = (time²)

900 s² = time²

Square root each side:

<em>T = 30 seconds</em>

7 0
3 years ago
You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

The minumum speed the pail must have at its highest point if no water is to spill from it

= 2.64 m/s

Explanation:

Working with the forces acting on the water in the pail at any point.

The weight of water is always directed downwards.

The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.

And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.

At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.

Net force = W + (normal force)

But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

Normal force = 0

Net force = W

Net force = centripetal force = (mv²/r)

W = mg

(mv²/r) = mg

r = 0.710 m

g = 9.8 m/s²

v² = gr = 9.8 × 0.71 = 6.958

v = √(6.958) = 2.64 m/s

Hope this Helps!!!

7 0
3 years ago
A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha
Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Where,m_{1} = mass of hunter

m_{2} = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

0+0=70\times v_{1}+0.042\times590

v_{1}=-\dfrac{0.042\times590}{70}

v=-0.354\ m/s

Hence, The recoil velocity is 0.354 m/s.

8 0
3 years ago
A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed
malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here:  brainly.com/question/25573309

#SPJ10

5 0
2 years ago
Describe the ratio of kinetic energy and potential energy the skateboarder will have at each of the following points in her ride
amm1812

Answer:

no answer

Explanation:

7 0
3 years ago
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