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2 years ago

3. What is the acceleration of a 10 kg mass pushed by a 5 N force?

Physics

2 answers:

Lesechka [4]2 years ago

7 0

Answer:

$\frac{1}{2} \frac{m}{s^2}$

Explanation:

Recall the equation F = ma. We want to find a, and we already know F and m, so we can plug those in. $5N = 10kg \cdot a -> a = 5N/10kg = \frac{1}{2} \frac{m}{s^2}$

insens350 [35]2 years ago

5 0

Answer:

F=ma

Plug it in:

5=10a

5/10=(10a)/10

.5m/s²=a

Explanation:

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Which use combustion to release thermal energy?

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2 years ago

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A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

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3 years ago

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3 years ago

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A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in

butalik [34]

Answer:

copper will have more change in temperature as compare with aluminum

Explanation:

Hot piece of copper is made in contact with cold piece of aluminium

So here thermal energy transfer will take place from copper to aluminium

so by energy conservation we can say that heat given by copper is same as the heat absorbed by aluminium.

now we have

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

here we know that

$s_1$ = specific heat capacity of copper

$s_2$ = specific heat capacity of aluminum

given that specific heat capacity of aluminium is more than double that of copper

so we can say

$s_2 = 2s_1$

so here if the mass of copper and aluminium is same then

$\Delta T_1 = 2 \Delta T_2$

so temperature change of copper is twice the temperature change of aluminium

So copper will have more change in temperature as compare with aluminum

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3 years ago