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FinnZ [79.3K]
2 years ago
14

I need help ASAP I need to get this right plz plz plz!!!!!

Physics
1 answer:
zhenek [66]2 years ago
3 0

Answer:

option d and b..............

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How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al
dusya [7]

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

Q = ms\Delta T

here we know that

m = 50 g

\Delta T = 110 - 10

\Delta T = 100 ^oC

so we have

Q = 50(0.95)(100)

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3 years ago
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A car has the velocity of 2.35 after 4.67 it's velocity is 9.89 what is the average acceleration
kiruha [24]
Average acceleration  =  (change in speed) / (time for the change) .
  
Change in speed = (ending speed) - (beginning speed)

                       =  (9.89 miles/hour) - (2.35 yards/second)  = 26,839.2 ft/hr

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3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
2 years ago
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