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Lemur
4 months ago
8

a carpenter hits a nail with a hammer. compared to the magnitude of the force the hammer exerts on the nail, the magnitude of th

e force the nail exerts on the hammer during contact is?
Physics
2 answers:
ValentinkaMS [3.4K]4 months ago
8 0

Final Answer:

Equal (and in the opposite direction)

Clarification:

According to Newton's third law of motion:

"When object A applies a force to object B, object B responds with an equal and opposite force back on object A. These forces are referred to as action and reaction."

In this scenario, the hammer applies force to the nail—if we consider the hammer as object A and the nail as object B—then in accordance with Newton's third law, the nail applies a force back on the hammer that is equal and opposite.

Thus, the force the nail applies to the hammer is equal to the force the hammer exerts on the nail (but in the opposite direction).

Sav [3.1K]4 months ago
7 0
Equal force will be applied, but in opposing directions.
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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

It is known that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2..............1

let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

6 0
3 months ago
Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w
Keith_Richards [3271]

Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

      U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    Em_{f} = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

    Em₀ = m 6.48

    Em_{f} = m 9.8 × 0.2

    Em_{f} = m 1.96

Thus the energy lost is given by:

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

8 0
3 months ago
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