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Ad libitum [116K]
3 years ago
9

Calculate the acceleration of a student riding his bicycle in a straight line that speeds up from 4 m/s to 6 m/s in 5 seconds.

Physics
1 answer:
solong [7]3 years ago
7 0
Acceleration is equal to velocity final minus velocity initial divided by time. 6m/s minus 4m/s divided by 5 seconds is 0.4m/s^2.

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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

4 0
3 years ago
Help ASAP, will give you brainliest!
Stels [109]

Answer:

A

Explanation:

5 0
3 years ago
Read 2 more answers
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis
tangare [24]

Answer:

56.86153 N

Explanation:

t =Time taken

F = Force

Power

P=\frac{W}{t}\\\Rightarrow W=P\times t\\\Rightarrow W=77\times 0.96\\\Rightarrow W=73.92\ Joules

Work done

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N

The magnitude of the force that is exerted on the handle is 56.86153 N

5 0
3 years ago
A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

            Force =  Spring constant x Elongation

            F = kx

Here force is weight of mass

           F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x  = 36 cm = 0.36 m

Substituting

           F = kx

           5.3 = k x 0.36

             k = 14.72 N/m

Spring constant is 14.72 N/m

6 0
2 years ago
If the original pressure is 5 atm and original volume is 100 L, and the new volume is 20 L, what is the new pressure?
Nana76 [90]
P1V1=P2V2
5*100=P2*20
500=20P
P=25
7 0
3 years ago
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