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aleksklad [387]
3 years ago
13

A book is pushed across the table by a student. The student stops pushing on the book and the book slows down and stops. The for

ce stopping the book is a force known as
A) inertia.
B) friction.
C) the applied force.
D) the action-reaction force.
Physics
2 answers:
V125BC [204]3 years ago
6 0
B. Friction. If the book was touching nothing, there would be no force to change its velocity given to it by the initial push. Because the book is sliding across the table, the molecules in the table are "slowing down" the book because they are exerting the force of friction on the book. What happens is that the charges in the book's molecules are attracted to the charges in the table's molecules. The attracting force is in the negative direction, causing the book to experience friction and therefore slow down.
VashaNatasha [74]3 years ago
3 0
A. inertia 

Because it has to do with the motion of something, especially if it changes its pace. In this example, the book's motion, when sliding on the table, decreased because of less force being given off from the student. 
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dem82 [27]
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There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:

(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2

We can reduce this to:

(15 x 10)m/36 s^2 = 150 m/36 s^2

Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
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3 years ago
Kinetic friction acts on a baseball player sliding into first base. Will the player's velocity change?
Law Incorporation [45]
Yes, his velocity will decrease the further he slides.
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3 years ago
A passenger in a helicopter traveling upwards at 15 m/s accidentally drops a package out the window. If it takes 15 seconds to r
Alexeev081 [22]

Answer:

The helicopter was 1103.63 meters high when the package was dropped.

Explanation:

We consider positive speed as a downward movement

y: height (m)

t: time (s)

v₀: initial speed (m/s)

Δy = v₀t + \frac{1}{2}gt²

Δy= 15\frac{m}{s}×15 s + \frac{1}{2}×9.81\frac{m}{s^{2} }×(15 s)²

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3 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

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