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2 years ago

Phys-1A Horizontal Practice 2

Physics

1 answer:

krek1111 [17]2 years ago

7 0

So basically you, then, finally, you

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A sled is slowing down at the bottom of a snowy hill.

Kitty [74]

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

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3 years ago

Find the magnitude of acceleration (ft/s^2) a person experiences when he or she is texting and driving 58mph, hits a wall, and c

SVEN [57.7K]

Answer:

350 ft/s²

Explanation:

First, convert mph to ft/s.

58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s

Given:

v₀ = 85.1 ft/s

v = 0 ft/s

t = 0.24 s

Find: a

v = at + v₀

a = (v − v₀) / t

a = (0 ft/s − 85.1 ft/s) / 0.24 s

a = -354 ft/s²

Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².

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3 years ago

What is the mass of an object which has a force of 600 N acting on it and is travelling

Paha777 [63]

Answer:

<em>The mass of the object is 40 Kg</em>

Explanation:

<u>Net Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object is:

F = m.a

Where:

a = acceleration of the object.

m = mass of the object.

The mass can be calculated by solving for m:

$\displaystyle m=\frac{F}{a}$

The object has a net force of F=600 N acting on it and travels at $a=15\ m/s^2$ , thus the mas is:

$\displaystyle m=\frac{600}{15}$

m = 40 Kg

The mass of the object is 40 Kg

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3 years ago

The diagram shows a stone suspended under the surface of a liquid from a string. The stone experiences a pressure caused by the

Ivan

Where is the diagram? What is the question?

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3 years ago

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c

bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

$U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.$

Given an arbitrary $x$ and its double, $2x$ , launch velocities are

$v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.$

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3 years ago