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3 years ago

Answer these questions please and you will get the brain list

Physics

1 answer:

Naddika [18.5K]3 years ago

7 0

I am using the equation F=ma (force equals mass times acceleration) to solve these problems.

1. You are looking for force, and have mass and acceleration. You just plug in the values for mass and acceleration to get the force needed.

F=(15kg)(5m/s^2)

F=75N

2. Again, you are looking for force, and just need to plug in the values for mass and acceleration

F=(3kg)(2.4m/s^2)

F=7.2N

3. In this problem, you have force and mass, but need to find acceleration. To do this, you need to get acceleration alone on one side of the equation - divide each side by m. Your equation will now be F/m=a

a=(5N)/(3.7kg)

a=18.5m/s^2

I did not use significant figures. Let me know if you need to do that and need any help on that. Hope this helps!

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

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3 years ago

An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit

lora16 [44]

Answer: The velocity of released alpha particle is $1.127\times 10^7m/s$

Explanation:

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

$m_1v_1=m_2v_2$

where,

$m_1\text{ and }v_1$ = Initial mass and velocity

$m_2\text{ and }v_2$ = Final mass and velocity

We are given:

$m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s$

Putting values in above equation, we get:

$238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s$

Hence, the velocity of released alpha particle is $1.127\times 10^7m/s$

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3 years ago

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