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s2008m [1.1K]
2 years ago
14

A force of 36N acts at an angle of 55 degrees to the vertical. The force moves its point of application by 64 cm in the directio

n of the force calculate work done by:
(A)the horizontal component of force


(B)the vertical component of the force
Physics
1 answer:
anastassius [24]2 years ago
7 0

The work done by the horizontal component of force is 18.89 Joules while the vertical component of the force is 13.22 Joules.

<h3>WORK </h3>

The work done on an object, is the product of the force F and the distance S in the direction of the force. That is,

W = F x S

The S.I unit of work is Joule

Given that a force of 36N acts at an angle of 55 degrees to the vertical. The force moves its point of application by 64 cm in the direction of the force

To calculate work done by:

(A) the horizontal component of force;

Force F = 36Sin55 = 29.489N

W = 29.489 x 64/100

W = 18.87 J

(B) the vertical component of the force;

Force F = 36Cos55 = 20.649

W = 20.649 x 64/100

W = 20.649 x 0.64

W = 13.22 J

Therefore, the work done by the horizontal component of force is 18.89 Joules while the vertical component of the force is 13.22 Joules.

Learn more about Work here: brainly.com/question/8119756

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A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

<u>v0 = 10 m/s</u>

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

<u>y = 1.2 m</u>

The maximum height above the ground is 1.2 m.

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What happens to most of the light waves that strike a clear pane of glass
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It’s C


Because C is a reflection which reflects something such as mirror

Hope this helps! •~•
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PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!
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<u><em>So here best appropriate graph must be option A</em></u>

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