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irina [24]
3 years ago
5

A 10-meter rope is lying on the floor and has a mass force of 20 N. How much work is required to raise one end of the rope to a

height of 5 meters?
Physics
1 answer:
VMariaS [17]3 years ago
7 0
Answer:10N

Explanation: I think
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A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?
lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee  = 5kg

Final speed  = 12m/s

Unknown:

Impulse of the frisbee  = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

 Impulse  = mass (Final velocity  -  Initial velocity)

 Impulse  = 5(12  - 0)  = 60kgm/s

3 0
3 years ago
What is a force, motion and energy??​
Leviafan [203]

Answer:Kinetic energy is the energy of motion. All moving objects have kinetic energy. When an object is in motion, it changes its position by moving in a direction: up, down, forward, or backward. 3.  A force is a push or pull that causes an object to move, change direction, change speed, or stop.

Explanation: Not sure if that's what you meant but that's the answer I can give you.

5 0
3 years ago
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Which type radiation can be observed well from Earth's surface?
umka2103 [35]

Answer:

Eletromagnetic radiation which is also known as visible light.

Explanation:

4 0
3 years ago
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What type of image results when the object is located between 2f (2 focal lengths) and f (the focal point) of a convex lens?
VikaD [51]

Answer:

C

Explanation:

the image formed is real but inverted and magnified

you can remember it by R I M

hope this helps

4 0
3 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
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