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Rom4ik [11]
3 years ago
9

A 12-kg hammer strikes a nail at a velocity of and comes to rest in a time interval of 8.0 ms. (a) What is the impulse given to

the nail
Physics
1 answer:
koban [17]3 years ago
5 0

Answer:

Impulse = 0.9408Ns

Explanation:

Impulse is the product of force and time

Impulse = Ft

Given

Force F = mg

F = 12(9.8)

F = 117.6N

Time = 8 * 10^-3 = 0.008s

Get the impulse

Impulse = 117.6*0.008

Impulse = 0.9408Ns

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An object accelerates 3 m/s2 , when a force of 6 N acts on it. What is the object’s mass
telo118 [61]

Answer:

<h2>2 kg</h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{6}{3}  \\  = 2

We have the final answer as

<h3>2 kg</h3>

Hope this helps you

3 0
2 years ago
80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
vodka [1.7K]

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

Therefore, the resulting temperature is 23.37 ⁰C

6 0
3 years ago
A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h
katen-ka-za [31]

Answer:

v_r=5.89\ m.s^{-1}

Explanation:

Given:

  • mass of rocket, m_r=50\ g
  • time of observation, t=2\ s
  • mass lost by the rocket by expulsion of air, m_a=10\%\ of m_r=5\ g
  • velocity of air, v_a=53\ m.s^{-1}

<u>Now the momentum of air will be equal to the momentum of rocket in the opposite direction: </u>(Using the theory of elastic collision)

m_a.v_a=(m_r-m_a)\times v_r

5\times 53=(50-5)\times v_r

v_r=5.89\ m.s^{-1}

7 0
3 years ago
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