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Juli2301 [7.4K]
4 years ago
10

Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin

g when determining the work done on the object?
Only the component of the force perpendicular to the motion is used to calculate the work.
If the force acts in the same direction as the motion, then no work is done.
When there is an angle between the two directions, the cosine of the angle must be considered.
A force at a right angle to the motion requires the use of the sine of the angle.
Physics
2 answers:
Zigmanuir [339]4 years ago
7 0

Answer:

C

Explanation:

erastova [34]4 years ago
4 0

Answer

(C).

When there is an angle between the two directions, the cosine of the angle must be considered.


Step by step Solution

The work done by a force is defined as the product of the force and the distance traveled in the direction of motion.

The first answer "Only the component of the force perpendicular to the motion is used to calculate the work" is wrong because, the force perpendicular to motion does no work.

The second choice "If the force acts in the same direction as the motion, then no work is done" is wrong because the work in the direction of the force is W=F\times d.

Fourth answer "A force at a right angle to the motion requires the use of the sine of the angle" is wrong because the sin(90)=0 meaning that there is no work done in the direction perpendicular to the motion.

The third answer" When there is an angle between the two directions, the cosine of the angle must be considered." is correct because the work is calculated using the force in the direction of the motion. The magnitude of this force is F\times d\times \cos(\theta).




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The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0
avanturin [10]

Answer:

(a) v(t) = 6t^2 - 6t - 12, a(t) = 12t - 6

(b) When 0 \leq t < 0.5, object is slowing down, when t > 0.5 object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

v(t) = \frac{ds(t)}{dt}  = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12

To get the acceleration function, we need to take the derivative of the velocity function.

a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

12t - 6 < 0 \\12t < 6 \\t < 0.5

On the other hand, object is speeding up when a > 0

12t - 6 > 0 \\12t > 6 \\t > 0.5

Therefore, when 0 \leq t < 0.5, object is slowing down, when t > 0.5 object is speeding up.

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4 years ago
One student bangs two bricks together
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Answer:

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Explanation:

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Answer:

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What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

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Answer:

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Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

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F_x = F cos \theta

with \theta=41.9^{\circ}

- The frictional force, whose magnitude is

F_f = \mu mg

where \mu=0.33, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N

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