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statuscvo [17]
3 years ago
12

Can someone please answer this for me

Physics
1 answer:
UNO [17]3 years ago
5 0

C

Because in my opinion they do bending of a light wave as it passes at an angle from one medium to another.

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In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller i
kotykmax [81]

The maximum angular speed of the roller is 3.47rad/a,  

the maximum tangential speed of the point an the rim of the roller is 1.47m/s,

the time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation is t = 2.78s,

5.4 rotation has the roller turned between t=0 and the time found in part c

<h3>What does tangential speed refer to?</h3>

Any item moving along a circular path has a linear component to its speed called tangential velocity. An object's velocity is always pointed tangentially when it travels in a circle at a distance r from the center. The word for this is tangential velocity.

To calculate the tangential speed, divide the circumference by the time required to complete one spin. For instance, if it takes 12 seconds to complete one rotation, divide 18.84 by 12 to get 1.57 feet per second as the tangential velocity.

(a) angular speed w = dθ/dt = 5t - 1.8t^2

dw/dt = 5 - 3.6t = 0 for max w

so max w occurs at t = 5/3.6 s = 1.39s

so w max = 5*1.39 - 1.8*(1.39)^2 = 3.47 rad/s

(b) tangential speed v = r*w

r = D/2 = 0.5m

so v = 0.5*w = 1.74 m/s

(c) w is positive until 5t = 1.8t^2

so t = 5/1.8s = 2.78s (or t = 0 invalid)

After t = 2.87s, w is negative (starts reversing direction of rotation)

Driving force would actually have to be removed some time before t=2.78s because the roller can't stop instantaneously, but insufficient info to calculate this.

(d) Up to t = 2.78s, θ = 2.5*(2.78)^2 - 0.6*(2.78)^3 rad = 33.95 rad = 5.40 rotations

Therefore,

A) 3.47rad/a

B) 1.47m/s

C) t = 2.78s

D) 5.4 rotation

To learn more about tangential speed, refer to:

brainly.com/question/19660334

#SPJ4

4 0
2 years ago
A ball, with a mass of 5.9kg, is thrown directly upwards. It reaches a maximum height of 10m from the point at which it was rele
katrin2010 [14]

Answer:

14 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, when the ball is thrown from the ground, it has only kinetic energy, which is given by

K=\frac{1}{2}mv^2

where m = 5.9 kg is the mass of the ball and v is its initial speed.

As the ball goes up, its speed decreases, so its kinetic energy decreases and converts into gravitational potential energy. When the ball reaches its maximum height, the speed has become zero, and all the kinetic energy has been converted into gravitational potential energy, given by:

U=mgh

where g = 9.8 m/s^2 is the gravitational acceleration and h = 10 m is the maximum height reached by the ball.

Since we can ignore air resistance, energy must be conserved, so the initial kinetic energy must be equal to the final potential energy of the ball, so we can write:

K=U\\\frac{1}{2}mv^2=mgh

And we can solve the equation to find v, the initial speed of the ball:

v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s


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