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Firdavs [7]
3 years ago
14

How is a quarter able to role on its edge?

Physics
1 answer:
Olegator [25]3 years ago
5 0
Idk its really weird because the ridges are making the quarter stand and therefore able to roll
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A wire that is 0.86 meters long is moved perpendicularly through a constant magnetic field of strength 0.035 newtons/amp·meter a
avanturin [10]
<span>The answer to your question is choice: D</span>
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2 years ago
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A 1200.0-kg car is traveling at 19m/s. The driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic f
Alja [10]
<h2>Answer</h2>

option D)

2.4 seconds

<h2>Explanation</h2>

Given in the question,

mass of car = 1200kg

speed of car = 19m/s

Force due to direction of travel

F = ma

  = 12000(a)

Force to due frictional force in reverse direction

-F = mg(friction coefficient)

   = -12000(9.81)(0.8)

<h2>-mg(friction coefficient) = ma  </h2>

(cancelling mass from both side of equation)

g(0.8) = a

(9.81)(0.8) = a

a = 7.848 m/s²

<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>

where vf = final velocity

          vo = initial velocity

          a = acceleration

           t = time

0 - 19 = 7.8(t)

t = 19/7.8

  = 2.436 s

  ≈ 2.4s

5 0
3 years ago
A ball is thrown upward with an initial velocity of +9.8 m/s. How high does it reach before it starts descending?
aksik [14]
Hope this helps you.

3 0
3 years ago
Suppose that two objects attract each other with a gravitational force of 32 Newtons. If the distance between the two objects is
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Answer:

64N

Explanation:

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5 0
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A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density
blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

6 0
3 years ago
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