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2 years ago

15

an object starts from rest with a constant acceleration of 8.0m/s^2 along a straight line.find the average speed for 5 second in

terval

1 answer:

frozen [14]2 years ago

6 0

Answer:

40 m/s

Explanation:

If you multiply the acceleration with time you get the average speed.

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You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t

uranmaximum [27]

Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together. This is done because the friction force is going to have to be compensated for. We will need that much more force than we otherwise would to achieve the desired acceleration:

$F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2} =0.0702N$

The friction force will be given by the normal force times the coefficient of friction. Here the normal force is just its weight, mg

$F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N$

Now the total force required is:

0.0702N+0.803N=0.873N

5 0

2 years ago

A man on a bicycle of total mass of 10kg has a Velocity of 2 ms.' He Paddles faster for 5 seconds and the velocity Increase to i

sesenic [268]

Answer:

the value of force, F=4.0N

Explanation:

Firstly, recall velocity-time equation

v=u+at
(4)=(2)+a(5)
a=0.4m/s²

Secondly, recall the Newton's 2nd Law

<em>F</em><em>=</em><em>ma</em>
<em>F</em><em>=</em><em>(</em><em>1</em><em>0</em><em>)</em><em>(</em><em>0</em><em>.</em><em>4</em><em>)</em>
<em>F</em><em>=</em><em>4</em><em>.</em><em>0</em><em>N</em>

7 0

1 year ago

If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return?

iren [92.7K]

Same speed, because mass is neglected. The things that affect the speed are the distance and speed of the rock.

5 0

3 years ago

How does the speed of the different molecules that make up the air depend on their masses?

koban [17]

Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?

8 0

2 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago