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3 years ago

What is the SI unit for graviational potential energy

Physics

1 answer:

scZoUnD [109]3 years ago

3 0

Gravitational potential energy is energy.
The unit of energy is the Joule.

1 Joule = 1 kilogram-meter² / sec²

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A spherical balloon is being inflated and the radius of theballoon is increasing at a rate of 2 cm/s.(A) Express the radius (r)

goblinko [34]

Answer:

A.) r = 2t

B.) V = 33.5t^3

Explanation:

Given that a spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cm/s

A) Express the radius (r) of the balloon as a function of the time (t).

Since the rate = 2 cm/s that is,

Rate = radius/ time

Therefore,

2 = r/t

Make r the subject of formula

r = 2t

(B) If V is the volume of the balloon as a function of the radius, find V or and interpret it.

Let assume that the balloon is spherical. Volume of a sphere is;

V = 4/3πr^3

Substitute r = 2t into the formula

V = 4/3π(2t)^3

V = 4/3π × 8t^3

V = 32/3 × πt^3

V = 33.5t^3

6 0

3 years ago

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to

SpyIntel [72]

This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s

3 0

3 years ago

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Anna [14]

Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

7 0

3 years ago

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A body of mass 200gram is executing SHM with amplitude of 20mm. The magnitude of maximum force which acts upon it is 0.6N. Calcu

igor_vitrenko [27]

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

$\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2$

Hence,the value of maximum velocity will be 0.3464 m/s².

To learn more about the velocity, refer to the link;

brainly.com/question/862972

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7 0

2 years ago

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Students hypothesized that by running an electric current through the wire of the apparatus shown here, they could cause a non-m

astra-53 [7]

The answer is d i think

5 0

3 years ago

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