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3 years ago

A force of gravity pulls downward on a book on a table. What force prevents the book from accelerating downward?

Physics

2 answers:

Nadusha1986 [10]3 years ago

3 0

Answer: The Normal Force

Explanation:

According to Third Newton's Law, also konwn as The principle of action and reaction :

"If a body A exerts an action on another body B, it performs on A another action that is the same and in the opposite direction."

In this case this is what happens with the book on a table, gravity force pulls it downward, but the Normal Force (perpendicular to the surface of the table) acts in the opposite direction with the same magnitude, preventing the book from accelerating downward.

Gnoma [55]3 years ago

3 0

Answer: The support / normal force

Explanation: The force that prevents downward acceleration is the support force.

The force that acts on the book at rest balancing up the gravitational force is the normal force.

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Post Test: Forces and Motion

Len [333]

Answer: B to C

Explanation: The line is curving inwards, practically calculating the stance that it had went down. If it went straight across, it stayed the same till a specific point, furthermore calculating the bent line bending upwards is actually a partial-raise, conclude points B to C is most likely an un-even balance, meaning it had went down; or decreasing. B to C is the decreasing segment of this equation/problem (question).

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3 years ago

A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i

laiz [17]

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==> x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==> v ≈ 1.43 m/s

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

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2 years ago

ASE (National Institute for Automotive Service Excellence) certified professionals must be retested every _______ years to keep

Bess [88]

Hi,

The correct answer is C. five

I hope I helped :)

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3 years ago

Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for

Zanzabum

Calm, sunny days with wind moving away from the center.

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3 years ago

Read 2 more answers

We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e

Tju [1.3M]

The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>

So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
Since the Moon and Sun are on opposite sides of Earth during a full moon, the light's distance traveled equals,

$d=(1.5*10^8km)+2(3.8*10^5km)=1.51*10^8km=1.51*10^{11}m$

As we know that light travels at a speed of 300,000 km per second. then, the time taken by the light reflected from sun to reach on earth will be,

$t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min$

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

Learn more about distance here:

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2 years ago