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3 years ago

What is the mass of a bird if it has a potential energy of 147J when flying at a height of 30 m above the ground?

Physics

1 answer:

Artist 52 [7]3 years ago

5 0

Answer:

0.5kg

Explanation:

Given parameters:

Potential energy = 147J

Height = 30m

Unknown:

Mass of the bird = ?

Solution:

Potential energy is the energy due to the position of a body. Now, the expression for finding the potential energy is given as;

P.E = mgH

m is the mass

g is the acceleration due to gravity = 9.8m/s²

H is the height

147 = m x 9.8 x 30

m = 0.5kg

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3 years ago

8. A crate of bananas weighing 3000 N is shipped from South America to New York, where it

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3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

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Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

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Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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3 years ago