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MArishka [77]
3 years ago
14

How does the energy possessed by a ball bearing change as it travels along an incline ramp?

Physics
1 answer:
ruslelena [56]3 years ago
7 0
Kinetic energy increases, potencial energy decreases,
kinetic energy + potential energy = energy, energy can not be destroyed, just transformed


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In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
2 years ago
On a popular amusement park ride, the riders stand along the inside wall of a large cylinder with a diameter of 8.27 m. The cyli
____ [38]

Answer:

Explanation:

D = 8.27 m   ⇒  R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W  ⇒  μ*N = m*g   <em>(I)</em>

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation <em>I</em>

<em />

μ*N = m*g    ⇒    μ*m*ω²*R = m*g   ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

3 0
3 years ago
How to reduce friction of moving a heavy load .
Vilka [71]
● Use lubricants such as oil.
● Place the load on a piece of cloth as it can ease the friction
● Place the load on a cart with wheels so it can be converted to rolling friction and can be pushed easily.
● Place the load on an inclined plane
3 0
3 years ago
A potter’s wheel moves from rest to an angular speed of 0.40 rev/s in 37.5 s.
svlad2 [7]

The angular acceleration of the potter's wheel is 0.067 rad/s².

The given parameters:

  • Final angular speed, ωf = 0.4 rev/s
  • Time of motion, t = 37.5 s

<h3>What is angular acceleration?</h3>
  • Angular acceleration of an object is the rate of change of angular speed of the object.

The angular acceleration of the potter's wheel is calculated as follows;

\alpha = \frac{\Delta \omega }{t} \\\\\alpha = (0.4\ rev/s \times \frac{2 \pi \ rad}{1 \ rev} ) \times \frac{1}{37.5 \ s} \\\\\alpha = 0.067 \ rad/s^2

Thus, the angular acceleration of the potter's wheel is 0.067 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

5 0
2 years ago
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi
maria [59]

Answer:

a) a = 5.03x10¹³ m/s²

b) V_{f} = 4.4 \cdot 10^{5} m/s

Explanation:    

a) The acceleration of the positron can be found as follows:

F = q*E    (1)

Also,

F = ma    (2)

By entering equation (1) into (2), we have:

a = \frac{F}{m} = \frac{qE}{m}

<u>Where:</u>

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C    

E: is the electric field = 286 N/C

a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2}

b) The positron's speed can be calculated using the following equation:

V_{f} = V_{0} + at

<u>Where</u>:

V_{f}: is the final speed =?

V_{0}: is the initial speed =0

t: is the time = 8.70x10⁻⁹ s

V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s

I hope it helps you!

4 0
2 years ago
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