All categories

3 years ago

Can anybody answer this its a science question NO LINKS !!!!

Physics

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

1) Potential energy (gravitation)

2)Potential energy

3)Kinetic energy

4) Kinetic energy

5)Potential energy

6)Kinectic energy

7)Potential energy (elastic)

Hope it helps

You might be interested in

A 50 kg object traveling at 100 m/s collides (perfectly elastic) with a 50 kg object initially at rest.

tino4ka555 [31]

Answer:

Explanation:

8 0

3 years ago

Immediately after it has started to rain, dust particles and oil residue float on the water and ____________ the traction of tir

Mkey [24]

When it rains, dust particles and oil residues float on the water and this reduces the traction of tires.

<h3>What is traction?</h3>

This concept refers to a force between the tires and road that causes the movement of the wheels or vehicle is slower.

<h3>What happens with traction when it rains?</h3>

It is well-known more accidents occur when it rains, which is caused by cars slidding on the road. This is because when it rains traction or the grip of the wheel drastically reduces.

Learn more about traction in: brainly.com/question/14525337

8 0

2 years ago

What is the buoyant force of a dog displacing 10 pounds of water?

scZoUnD [109]

Answer:

find V using d = M/V

Explanation:

F = d*V*g

d = density of fluid (in this case, 1000)

V = volume of object

g = gravity

3 0

3 years ago

Please help me with my physics! I do not understand and please provide work.

Marrrta [24]

26) A) 400J as GPE = mgh
27) C) Both obj has same PE
28) B) It decreases as energy is not being provided.
29) C) (42 - 0)/7 = 6m/s2

5 0

3 years ago

Read 2 more answers

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