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gizmo_the_mogwai [7]
4 years ago
14

A hockey player uses her hockey stick to exert a force of 7.957.95 N on a stationary hockey puck, which is resting on a nearly f

rictionless ice surface. The hockey puck has a mass of 181181 g. What is the acceleration of the hockey puck?
Physics
1 answer:
Firlakuza [10]4 years ago
3 0

Answer:

44 m/s^2

Explanation:

Force, F = 7.95 N

mass, m = 181 g = 0.181 kg

By use of Newton's second law

F = m x a

a = F / m = 7.95 / 0.181 = 44 m/s^2

Thus, the acceleration of the hockey puck is 44 m/s^2.

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You are trying to overhear a juicy conversation, but from your distance of 23.0 m , it sounds like only an average whisper of 40
Len [333]
If its asking the distance for the 65 db then use a proportion, if otherwise pleas clarify. It sounds like a pretty juicy conversation.
4 0
4 years ago
The colour of star depend on its temperature, why?​
taurus [48]

<em>Another key factor that determines a star's colour is its temperature. As stars become hotter, the overall radiated energy increases, and the peak of the curve changes to shorter wavelengths. To put it another way, when a star heats up, the light it produces moves toward the blue end of the spectrum.</em>

4 0
3 years ago
A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s
pashok25 [27]

Answer:

109656.25 Nm

Explanation:

\omega_f = Final angular velocity = 1.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm

The torque specifications must be 109656.25 Nm

5 0
3 years ago
Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
4 years ago
During an eclipse the sun the earth and the moon act as a light source,an obstacle or screen=
Zigmanuir [339]

Answer:

Un eclipse solar se produce cuando la luna se interpone en el camino de la luz del sol y proyecta su sombra en la Tierra. Eso significa que durante el día, la luna se mueve por delante del sol y se pone oscuro. ... Este eclipse total se produce aproximadamente cada año y medio en algún lugar de la Tierra.

Explanation:

espero que te

haya

servido

6 0
3 years ago
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