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UNO [17]
2 years ago
6

Using the law of conservation of energy, what will be the KE of an arrow having a PE of 65J after it is shot from a bow?

Physics
1 answer:
Colt1911 [192]2 years ago
7 0

Answer:

KE = 65 J (The potential energy stored in the stretched bow is converted to kinetic energy of the moving bow)

Explanation:

Law of conservation of energy states that energy can neither be created nor be destroyed but transferred from one form to another. In other words, the total energy of the system at any point is always a constant.

Now, when the bow is in stretched position, the energy associated with the stretched bow is potential energy (PE).

The energy associated with motion of the body is kinetic energy (KE).

So, the stretched bow is not moving. So, KE at the start is 0.

Therefore, total energy initially is given as:

Initial total energy = PE + KE = PE + 0 = 65 J

Now, when the bow is shot, the bow is in motion. So, the total energy of the bow is only due to kinetic energy associated with it now. Also, the potential energy of a moving bow is 0.

Therefore, final total energy is given as:

Final total energy = Final KE + Final PE = Final KE + 0 = Final KE

Now, we know, from conservation of energy,

Initial total energy = Final total energy

⇒ 65 J = KE

or Final KE = 65 J

Therefore, the KE of an arrow having a PE of 65J after it is shot from a bow is 65 J.

Hence, the potential energy stored in the stretched bow is converted to kinetic energy of the moving bow.

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Sunspots are typically cooler than their surroundings. For example, the temperature of a large sunspot can be about 4,000 Kelvin which is lower than the temperature of the photosphere around it which is about 5,800 Kelvin.

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7 0
3 years ago
A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm
Marina86 [1]

Answer:

v(0)=2m/s

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

v(t)=\omega  *A*cos(\omega t + \phi)

Where:

\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase

Express the amplitude in meters:

10cm*\frac{1m}{100cm} =0.1m

The angular frequency can be found using the next equation:

\omega=\sqrt{\frac{k}{m} }

Using the data provided:

\omega=\sqrt{\frac{400}{1} } =20

At the equilibrium position:

\phi=0

v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s

6 0
3 years ago
A thin block of soft wood with a mass of 0.078 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is
natali 33 [55]

Answer:

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Explanation:

As we know that there is no external force on the system of wooden block and the bullet

so we can say momentum of the system is conserved here

so here we can say

P_i = P_f

m_1v_1 = m_1v_{1f} + m_2v_{2f}

4.67\times 10^{-3}(613) = 4.67 \times 10^{-3}v_{1f} + (0.078)(23)

so we will have

2.86 = 4.67\times 10^{-3} v_{1f} + 1.794

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Answer:

to me I think is static

Explanation:

that is my own thinking

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3 years ago
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