Which type of error is a false positive

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

The ball rotates 6.78 revolutions.

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

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

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is $252.52\ m/s^2$ .

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = $\frac{2500}{9}\ m/s$

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

$\therefore 0=\frac{2500}{9} + a(1.1)$

$\Rightarrow a=-\frac{2500}{9(1.1)}$

$\therefore a = - 252.52 \ m/s^2$

Hence , the deceleration of the rocket is $252.52\ m/s^2$ .

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0

2 years ago

A 2kg book is moved from a shelf that is 2m off the ground to a shelf that is 1.5m off the ground, what is it’s change in gravit

Ket [755]

The gravitational energy is going up subtracting the energy that was on the ground

3 0

4 years ago

Calculate the G.P.E. in joules of a 675-newton climber at the top of a 3,050-meter mountain in Colorado.

almond37 [142]

Answer:2673

Explanation:

4 0

3 years ago

Write any two features of capital

amid [387]

Washington DC and new Mexico

6 0

3 years ago

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