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adelina 88 [10]
2 years ago
7

Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds

Physics
1 answer:
solong [7]2 years ago
7 0

Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

=> 10a=11

=> a=11/10=1.1 m/(s)^2

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Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is
Llana [10]

Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

4 0
3 years ago
If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have?
Travka [436]

Answer:

Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.

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7 0
2 years ago
Which statement best describes beaches? Beaches change over time as waves move further up the shore. Beaches are always changing
NemiM [27]

Answer:

I think the 1st statement is right.

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