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Naya [18.7K]
3 years ago
6

You serve a volleyball with a mass of 2.5 kg. The ball leaves your hand with a speed of 23 m/s. What is the kinetic energy of th

e ball?
Physics
1 answer:
Zarrin [17]3 years ago
8 0

Answer:

661.25, I believe!

Explanation:

KE = (1/2)(2.5)(23)^2

KE=(1.25)(529)

KE= 661.25

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

5 0
3 years ago
A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo
user100 [1]

Answer:

\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

Explanation:

The total force on the particle is given by

\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}

Then, by replacing we have:

q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

where the cross product can be made with the determinant method.

Hope this helps!!

3 0
4 years ago
Read 2 more answers
On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar obje
Harrizon [31]

Answer: no u

Explanation: no u

8 0
3 years ago
A lightweight vertical spring of force constant k has its lower end mounted on a table. You compress the spring by a distance d,
shusha [124]

Answer:

v=d\sqrt{\frac{k}{m}}

Explanation:

In order to solve this problem, we can do an analysis of the energies involved in the system. Basically the addition of the initial potential energy of the spring and the kinetic energy of the mass should be the same as the addition of the final potential energy of the spring and the kinetic energy of the block. So we get the following equation:

U_{0}+K_{0}=U_{f}+K_{f}

In this case, since the block is moving from rest, the initial kinetic energy is zero. When the block loses contact with the spring, the final potential energy of the spring will be zero, so the equation simplifies to:

U_{0}=K_{f}

The initial potential energy of the spring is given by the equation:

U_{0}=\frac{1}{2}kd^{2}

the Kinetic energy of the block is then given by the equation:

K_{f}=\frac{1}{2}mv_{f}^{2}

so we can now set them both equal to each other, so we get:

=\frac{1}{2}kd^{2}=\frac{1}{2}mv_{f}^{2}

This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

kd^{2}=mv_{f}^{2}

so now we can solve this for the final velocity, so we get:

v=d\sqrt{\frac{k}{m}}

6 0
3 years ago
Based on their colors, which of the following stars is hottest? Which is coolest? Archenar (blue), Betelgeuse (red), Capella (ye
Cloud [144]

Answer: The hottest star is Archenar( blue) and the coolest star is Betelgeuse

Explanation:

Objects emit radiation that depends exclusively on their temperature. At an ambient temperature, the radiation emitted by an object is in the infrared spectrum (we could only see it with a special camera). If we heat it we will see that it first turns red (whose state we call “red hot”) because it is the lowest and least energetic wavelength of all.

If we continue to heat it, the wavelength that it emits to one with more energy will continue to increase and we will see that it turns yellow and then white. This is a signal that is emitting at all frequencies (but mainly in blue).

If we continue to warm a body that is "white hot", it would emit in the ultraviolet spectrum, with what would become ... black! then we would not see it emits light in the visible spectrum (well, we would see a very faint bluish light corresponding to the tail of the distribution of the spectrum it emits, but the peak of that spectrum would be in the ultraviolet).

6 0
3 years ago
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