All categories

3 years ago

Please help

Physics

1 answer:

garri49 [273]3 years ago

4 0

Answer:

Exerxingtransforms the chemical energy in the food you eat into Kinetic Energy.

Therefore, the answer is A. Kinetic Energy

You might be interested in

Dense fluids exert___________buoyant force then less dense fluids because they have more mass within a certain volume than a les

sergij07 [2.7K]

Answer:

The answer is 'more' as more mass can exert more pressure

6 0

3 years ago

Does air pressure increase or decrease with an increase in altitude?

Yakvenalex [24]

Answer:

increase

Explanation:

think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing

5 0

3 years ago

Read 2 more answers

Please help<br> question #70

Keith_Richards [23]

can i get the question so that i can answer your question

8 0

3 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

Read 2 more answers

Name and describe the apparatus used by Cavendish to discover the universal gravitation constant

Evgen [1.6K]

Its like a suspended wood with a lead sphere attached to each of its ends

3 0

4 years ago