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3 years ago

If a motorcycle starts from rest and maintains a constant acceleration of 3 m/s^2, what will its velocity be after 10 s?

Physics

2 answers:

Aneli [31]3 years ago

7 0

The answer is B. 30 meters per second

zvonat [6]3 years ago

5 0

In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

hope this helps XD

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3 years ago

With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas

xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

$\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity$

As it just touches the 15 mm high roof, the final velocity will be zero. So,

$v_f=0\ m/s$ .

Now, the change in kinetic energy is equal to:

$\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2$

Change in gravitational potential energy = Final PE - Initial PE

So,

$\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J$ [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

$0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s$

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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3 years ago

A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of

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A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.

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Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.

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The proton will be subject to a magnetic pull that is directed into the page.

Hence, option B is correct.

To learn more about the right-hand thumb rule refer to the link;

brainly.com/question/11521829

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