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alisha [4.7K]
4 years ago
5

What various forms of energy are present as you shoot a bow andarrow?

Physics
1 answer:
kodGreya [7K]4 years ago
8 0

Answer:

In order to shoot an arrow with a bow, you should hold the bow with your left hand, and should pull the string to yourself and stay for a moment to aim.

At that point your arm applies a force, and that force does work on the bow. This work is stored as elastic potential energy. When you release the arrow, this potential energy is converted to kinetic energy, in this way the arrow is thrown forward.

This is the simplified version. Of course, in real life the case is different.

In real life, when you hold the string to aim, there is elastic potential energy, and also gravitational potential energy to hold the arrow above the ground. When you release the arrow, it goes through the side of the bow with a friction. So, part of the potential energy is converted to heat when the arrow and bow is still in contact. After they lose contact, the potential energy is converted to kinetic energy, and the motion of the arrow is projectile, since the arrow  both moves forward and downwards. Eventually the arrow hits the ground.

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4 years ago
Calculate the energy needed to change 200.0 g of ice from -20.0°C to water at 35.0°C.
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3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
4 years ago
This illustration shows two opposing forces pulling on a wagon. Which description best describes how the wagon will move?
Talja [164]

Answer:

The wagon will move to the right.

Explanation:

From the question given above, the following data were obtained:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Direction of the wagon =.?

To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 30 – 10

Fₙ = 20 N to the right

From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.

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3 years ago
Use the right-hand rule for magnetic force to determine
Sphinxa [80]
The charge on the moving particle
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3 years ago
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