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3 years ago

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Gravitational Potential Energy: You and your friend, who weighs the same as you, want to go to the top of the Eiffel Tower. Your

friend takes the elevator straight up. You decide to walk up the spiral stairway, taking longer to do so. Compare the gravitational potential energy of you and your friend, after you both reach the top.

1 answer:

ololo11 [35]3 years ago

7 0

Answer:

Gravitational Potential Energy of both will be equal

Explanation:

The gravitational potential energy is given by the following formula:

$P.E = mgh$

where,

P.E = Gravitational Potential Energy

m = mass of object

g = acceleration due to gravity

h = height of object

Therefore, the gravitational potential energy of an object depends only upon the height of the object and its mass.

Since, both masses and the heights achieved by both people is equal. <u>Therefore, their gravitational potential energies will also be equal</u>

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Which describes the refraction of a wave?

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The answer is B. light changes direction moving through the lens of a magnifying glass

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3 years ago

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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

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

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

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

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

I need a creative science fair title about how music affects your run

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4 years ago

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a skier speeds along a flat patch of snow, and then flies horizontally off the edge at 16.0 m/s. He eventually lands on a straig

geniusboy [140]

Answer:

1.63 s

Explanation:

The skier lands on the sloped section when the direction of its velocity is exactly identical to that of the slope, so at $45.0^{\circ}$ below the horizontal.

This occurs when the magnitude of the vertical velocity is equal to the horizontal velocity (in fact, $\tan \theta =\frac{|v_y|}{v_x}$ , and since $\theta=45.0^{\circ}, tan \theta = 1$ and so $|v_y| = v_x$ .

We already know the horizontal velocity of the skier:

$v_x = 16.0 m/s$

And this is constant during the entire motion.

The vertical velocity instead is given by

$v_y = u_y + gt$

where

$u_y = 0$ is the initial vertical velocity (zero since the skier flies off horizontally)

g = 9.8 m/s^2

t is the time

Here we have chosen the downward direction as positive direction.

Substituting $v_y = 16.0 m/s$ , we find the time:

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3 years ago

A 6cm diameter horizontal pipe gradually narrows to 4cm.

Rom4ik [11]

Answer: $Q=0.5612 m^3/s$

Explanation:

Given

diameter of pipe( $d_1$ )=6 cm

diameter of pipe( $d_2$ )=4 cm

$P_1=32 kPa$

$P_2=24 kPa$

$A_1=\frac{\pi }{4}6^2=9\pi cm^2$

$A_2=\frac{\pi }{4}4^2=4\pi cm^2$

$v_1=\frac{Q}{A_1}$

Applying bernoulli's equation

$\frac{P_1}{\rho g}+\frac{v^2_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v^2_2}{2g}+z_2$

$\frac{P_1}{\rho g}+\frac{\frac{Q^2}{A_1^2}}{2g}+z_1=\frac{P_2}{\rho g}+\frac{\frac{Q^2}{A_2^2}}{2g}+z_2$

since $z_1=z_2$

$\frac{32\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_1^2g}=\frac{24\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_2^2g}$

$Q^2=\frac{8\times 2\times 81\pi ^2\times 16\pi ^2\times 10^{-4}}{65\pi ^2}$

$Q^2=3149.3722\times 10^{-4}$

$Q=\sqrt{3149.3722\times 10^{-4}}$

$Q=0.5612 m^3/s$

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3 years ago