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gulaghasi [49]
3 years ago
15

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Physics
1 answer:
Stolb23 [73]3 years ago
6 0

Answer:

200 cal/^{\circ}C

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

Q=C\Delta T

where

Q is the heat supplied

C is the heat capacity of the object

\Delta T is the change in temperature

In this problem we have:

Q=10,000 cal is the energy supplied

\Delta T=75C-25C=50C is the change in temperature of the object

Therefore, the heat capacity of the object is:

C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C

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Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

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This is the time of fall of the body to touch the ground

3 0
4 years ago
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

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Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

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3 0
3 years ago
Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.
Mars2501 [29]

Answer:

0.5\times 10^{-6}C

Explanation:

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We have given force F =0.036 N

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The acceleration goes up.
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Answer: This is what I found hope it helps

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