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Nesterboy [21]
2 years ago
12

When the 3.0 kg cylinder fell 500 m, the final temperature of the water was

Physics
1 answer:
miss Akunina [59]2 years ago
7 0

Answer:

Change in temperature =7.14° C

Explanation:

Assuming all potential energy of cylinder is converted into heat energy to increase the temperature .

Work done by gravity in first case =m *g*h= 3*10*500=15000

Let quantity of water be 1 kg at 0°C.

Also specific heat capacity of water is =4200 Joule

So potential energy is converted to m*4200*T(1) =15000

                      T(1)=3.57

Work done by gravity in first case =m *g*h= 9*10*500=45000

Here also potential energy is converted into m*4200*T(2)=45000

                    T(2)=10.71

Here temperature is increasing.

Thus change in temperature =T(1) -T(2) =7.14° C

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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

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Set up force equation

∑F=ma

∑F=W+FB

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2 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
2 years ago
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