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3 years ago

A 2.0 kg mass is lifted 4.0 m above the ground. find the change in gravitational energy

Physics

1 answer:

Makovka662 [10]3 years ago

6 0

Answer:

78.4 J

Explanation:

Given:

Mass of the object (m) = 2.0 kg

Height or displacement of the mass (h) = 4.0 m

Now, change in gravitational potential energy (ΔU) = ?

Change in gravitational potential energy of an object is the energy stored by the object when it is raised to some height above the reference ground level.

Change in gravitational potential energy of an object of mass 'm' raised to a height 'h' above the ground is given as:

$\Delta U=mgh$

Where, 'g' is the acceleration due to gravity and has a value of 9.8 m/s².

Now, plug in all the given values and solve for ΔU. This gives,

$\Delta U=(2.0\ kg)(9.8\ m/s^2)(4.0\ m)\\\\\Delta U=78.4\ J$

Therefore, the change in gravitational potential energy is 78.4 J.

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jek_recluse [69]

The approximate speed of sound is about 761 mph or 335.28 meters/second.

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2 years ago

n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

The potential of the soap bubble is $V_1 =307.0 V$

The new radius of the soap bubble is $R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m$

The initial electric potential is mathematically represented as

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

where k is the coulomb's constant with value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

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3 years ago

Why is moon is non-lomis back

qwelly [4]

Explanation:

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4 years ago

A trough is filled with a liquid of density 835 kg/m3. The ends of the trough are equilateral triangles with sides 4 m long and

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Answer:

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Explanation:

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volume=64 m^3

mass=835*64

mass=53440 Kg

Hydrostatic force=53440*9.8=523712 N

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4 years ago

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Marat540 [252]

B. fungi

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4 years ago