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3 years ago

2. A volleyball player strikes a 3 kg volleyball with a velocity of 14 m/s. What is the kinetic energy of the volleyball?

Physics

2 answers:

Effectus [21]3 years ago

6 0

Answer:

2. 294 j

3. 37500 J

Explanation:

Anarel [89]3 years ago

3 0

Answer:

2)294 3)15

Explanation:

2)E=mv^2/2

E=3×14^2/2

E=3×196/2

E=3×98

E=294 J

3)E=mv^2/2

m=2E/v^2

m=2×750/10^2

m=2×750/100

m=750/50

m=15

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A moving truck has more __ energy than a parked truck

Karo-lina-s [1.5K]

Kinetic energy than parked

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3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago

When measuring accelerations with three different graphs (x-t, v-t, a-t) which gives the most reliable results and why?

kipiarov [429]

Answer:

the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

Explanation:

The different graphics depending on time give various information, let's examine what we can get from some

Graph of x -t. from this graph we can obtain the speed through the slope, but the acceleration is not directly obtainable

v-t chart. We can get the acceleration not through the slope and the distance traveled by the area under the curve. Obtaining acceleration is very accurate since it is an average that avoids possible errors in measurements. This is the best graph to find the acceleration

Graph of a-t In this graph the acceleration is a point on the Y axis, it gives some errors because it depends strongly on the possible experimental errors.

In conclusion, the best graph to find the acceleration is v-t since calculating the slope averages the different experimental errors.

5 0

3 years ago

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by

∑F = ma (1)

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:

a_max = -w^2A (2)

3- The angular frequency w of a wave is related to the frequency f by:

w = 2π f (3)

Given Data

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston. </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle. </em>

Solution

(a)

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:

∑F_y -mg= ma

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So

0 = m(g + a)

a = —g

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):

a = —w^2A

where a = —g at the highest point. So

g = w^2A

Solve for w:

w =√g/A

Substitute for w from Equation (3):

2πf = √g/A

Solve for f :

f = 1/2π√g/A

Substitute numerical values:

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

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3 years ago

Which term describes the process of transferring charge without the direct contact

Arada [10]

Induction............

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3 years ago