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Xelga [282]
3 years ago
15

A book is sitting on a desk. what best describes the normal force acting on the book

Physics
2 answers:
Margarita [4]3 years ago
7 0

the upward force the desk pushes on the book.

from the free body diagram of the book, it is clear that on the book, there are two forces acting.

1) normal force F by the desk in Upward direction.

2) force of gravity by earth in downward direction.

The Upward normal force by the desk on the book balances the force of gravity by earth in downward direction. that is why the book remains stationary on the desk.

blsea [12.9K]3 years ago
4 0

The forces acting upon the book are shown below. The force of gravity pulling downward and the force of the table pushing upwards on the book are of equal magnitude and opposite directions. These two forces balance each other.

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KIM [24]

Answer rain gauge measures rain shadow units millimetres

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What happens if different lights are shined through a prism
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The light will bend when in 
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A block of mass 1.5 hangs at the of end of a weight cord suspended from the ceiling.what is the tension in the cord, and with wh
Len [333]

The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

<h3>What is the tension in the cord?</h3>

The tension in the cord is calculated as follows;

T = ma + mg

where;

  • a is the acceleration of the block
  • g is acceleration due to gravity
  • m is mass of the block

T = m(a + g)

T = 1.5(a + 9.8)

T = 1.5a + 14.7

Thus, the tension in the cord is (1.5a + 14.7) N.

If the block is at rest, the tension is 14.7 N.

<h3>Force of the force</h3>

The force with which the cord pulls is equal to the tension in the cord

F = T = m(a + g)

F = (1.5a + 14.7) N

If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.

Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.

Learn more about tension here: brainly.com/question/187404

#SPJ1

4 0
1 year ago
A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a d
Snowcat [4.5K]

Answer:

The energy stored is 1.4 x 10^-9 J.

Explanation:

Side of square, L = 10 cm = 0.1 m

Distance, d = 2 mm = 0.002 m

Electric field, E = 4000 V/m

The energy stored in the capacitor is

U = 0.5 C V^2

The capacitance is given by

C = \frac{\varepsilon o A}{d}\\\\So \\\\U = 0.5\frac{\varepsilon o A}{d}\times E^2 d^2\\\\U = 0.5\times 8.85\times 10^{-12}\times 0.1\times 0.1\times 4000\times 4000\times 0.002\\\\U = 1.4\times10^{-9} J

7 0
3 years ago
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