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3 years ago

10

The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans

verse direction radial line. normal direction.

1 answer:

Marianna [84]3 years ago

7 0

Tangent to the pathh.

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A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

timofeeve [1]

The solution would be like this for this specific problem:

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago

1.- An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct?

Nady [450]

Answer:

the correct one is C

Explanation:

For this exercise we must use the work definition

W = F. s

Where the bold characters indicate vectors and the point is the scalar producer

W = F s cos θ

Where θ is the angles between force and displacement.

Let us support this in our case. The cable creates an upward tension and with the elevator going down the angle between them is 180º so the work of the cable on the elevator is negative.

The evade has a downward force, its weight so the force goes down and the displacement goes down, as both are in the same direction the work is positive

When examining the statements the correct one is C

4 0

3 years ago

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

$a = g = 9.8 m/s^2$

Part b)

Now from kinematics equation we will have

$y = v_i t + \frac{1}{2} at^2$

now we have

y = 25 m

so from above equation

$25 = 0 + \frac{1}{2}(9.8 )t^2$

$t = 2.26 s$

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

$v_x = vcos\theta$

$v_x = 20 cos65 = 8.45 m/s$

$v_y = vsin\theta$

$v_y = 20 sin65 = 18.13 m/s$

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

$v_f - v_y = at$

$0 - 18.13 = (-9.8) t$

$t = 1.85 s$

so it will take 1.85 s to reach the top

5 0

3 years ago

The Graph above shows the speed of a car traveling in a straight line as a function of time. The car accelerates uniformly and r

devlian [24]

Answer:

Where is the graph??

If a car travels from zero to 4 m/s ins 8 sec

a = 4 / 8 = .5 m/s^2

V (2) = 2 * .5 = 1 m/s after 2 sec

S = V t + 1/2 a t^2

S = 1 * 3.9 + 1/2 * 1/2 *3.9^2 = 3.9 + 3.80 = 7.70 m from 2 to 5.9 sec

Check:

Total distance traveled = 1/2 a t^2 = 5.9^2 / 4 = 8.70 m

Distance traveled in 2 sec = 1/2 * 1/2 * 4 = 1 m

Total distance from 2 to 5.9 = 8.7 - 1 = 7.7 m agreeing with thw above

3 0

2 years ago

How often do industrial oxygen cylinders need to be tested

hjlf

Answer:

5 years

5 0

2 years ago