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3 years ago

When an object is thrown upwards and reaches its maximum height its speed is: a. Greater than the initial

Physics

1 answer:

suter [353]3 years ago

5 0

Answer:

Option d

Explanation:

When we throw an object in the upward direction, we provide it with certain initial velocity due to which it covers a certain distance up to the maximum height.

While the object is moving in the upward direction, its velocity keeps on reducing due to the acceleration due to gravity which acts vertically downwards in the opposite direction thus reducing its velocity.

So, the maximum height attained by the object is the point where this upward velocity of the body becomes zero and after that the object starts to fall down.

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What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?

sladkih [1.3K]

formula for gravitational P.E =mgh

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

7 0

3 years ago

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

We know that kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

m=mass

u=velocity

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 65\times 5.2^2\ J$

KE=878.8 J

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J$

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0

3 years ago

A machine does 1200 J of work in 1 min. What is the power developed

densk [106]

Answer:

20 watts

Explanation:

Big brain mode activated:

Power=1200J/60sec

Power=20 watts

8 0

3 years ago

A dog, that has a mass of 16 kgs, runs across a yard. What is the average force applied to the dog from the ground as it runs ac

VikaD [51]

Answer:

156.96 N

Explanation:

F=ma where m is the mass and a is acceleration

Substituting 16 Kg for m and 9.81 m/s2 for g then

F=16*9.81= 156.96 N

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3 years ago

Assume you built a really big machine that could launch the projectile a “significant” distance; for instance, several hundred m

Gwar [14]

<span>First, a problem would be the sheer amount of wind resistance. If an object travels as far as even just one hundred miles it could encounter different wind patterns that could change the trajectory of the object. Second would be the size of the projectile. This creates a problem because the bigger it is, the more momentum it could potentially pick up, given that it is not too big to complete the distance. This is another problem with size, how far the projectile can actually travel. You would have to actually calculate the ideal size of said object to make sure it could actually make the distance you're looking for. hope it helps:)</span>

3 0

3 years ago

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