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topjm [15]
3 years ago
7

A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How mu

ch work is done by this force as the object moves 6.0 m?
Physics
1 answer:
puteri [66]3 years ago
3 0

Answer:

84.6 J

Explanation:

The work done by the force is given by

W=Fd cos \theta

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

\theta=20^{\circ} is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J

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A pebble is dropped from rest from the top of a tall cliff and falls 53.4 m after 3.3 s has elapsed. How much farther does it dr
jeka94

Answer:

426.84 m

Explanation:

initial velocity u = 0

time t = 3.3 s

distance travelled s = 53.4 m

acceleration due to gravity = g

s = ut + 1/2 g t²

53.4 = 0 + 1/2 g x 3.3²

g = 9.8 m /s²

For the whole length of fall

distance travelled = h

total time = 6.6 + 3.3 = 9.9 s

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h = .5 x 9.8 x 9.9²

= 480.24 m

distance travelled in last 6.6 s

= 480.24 - 53.4

= 426.84 m

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2 years ago
The rotational speed of earth is similar to?​
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A swing has a period of 10 seconds. What is its frequency ?
kow [346]

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You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type
alexdok [17]

Answer:

30 N \cdot m

Explanation:

The torque applied by a force can be calculated as

\tau = F d sin \theta

where

F is the magnitude of the force

d is the length of the arm

\theta is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

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Substituting into the equation, we find

\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m

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3 years ago
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