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amid [387]
2 years ago
7

You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo

rce on the shopping cart ? b) What is the force of friction on the shopping cart ? c) What will happen to the shopping cart if you come across a wet portion of the parking lot and you continue to apply the same force ?
Physics
1 answer:
AlekseyPX2 years ago
4 0

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

Learn more about frictional force here: brainly.com/question/24386803

#SPJ1

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Explanation:

hope this helps have a good day

6 0
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A cycle track is 500 metres long. Amy completes 10 laps. She travelled at an average speed of 12.5 metres per second. She puts o
yanalaym [24]
The fast lap is irrelevant to the question, because it didn't happen
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To be perfectly technical about it, we don't actually have enough
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IF we can assume that Amy maintained a totally steady pace through
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5 0
3 years ago
Calculate the total surface area of cylinder whose base radius 2cm and height is 10cm​
murzikaleks [220]

Answer:

Answer :-  150.79cm2

Explanation:

                                               A=2πrh+2πr2

                                                 2·π·2·10+2·π·22

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The average depth of Indian Ocean is about 3000m.Calculate the fractional compression. ▲v/v, of water at the bottom of the ocean
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Answer:

∴ fractional compression = 1.34 × 10⁻²

Explanation:

given,

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We know,

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P₀ is the atmospheric pressure

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ρ is the density of the water, 1000 Kg/m³

P = 10⁵ + 1000 × 9.8 × 3000 = 2.94 × 10⁷ N/m²

using formula,

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B is bulk modulus and { -∆V/V} is the fractional compression

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\dfrac{-\Delta V}{V} =1.34 \times 10^{-2}

∴ fractional compression = 1.34 × 10⁻²

7 0
3 years ago
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