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densk [106]
1 year ago
11

A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote

ntial energy increases by 3773 J, determine the increase in potential energy of his mother if her mass is 56.0 kg.
Physics
1 answer:
notka56 [123]1 year ago
8 0

The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855#SPJ1

#SPJ1

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Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
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Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm
lina2011 [118]

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

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