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3 years ago

PLEASE HELP ME :)

Physics

1 answer:

ddd [48]3 years ago

8 0

A) 350 J

- The initial internal energy of the cup is

$U_i = 230 J$

- The final internal energy of the cup is

$U_f = 580 J$

According to the first law of thermodynamics:

$U_f - U_i = Q+W$

where

Q is the heat absorbed by the system

W is the work done on the system

The work done on the system in this case is 0, so we can rewrite the equation as

$U_f - U_i = Q$

And so we find the heat transferred

$Q=580 J - 230 J=350 J$

B) IN the cup

Explanation:

in this situation, we see that the internal energy of the cup increases. The internal energy of an object/substance is proportional to its temperature, so it is a measure of the average kinetic energy of the molecules of the object/substance. Therefore, in this case, the temperature (and the energy of the molecules of the substance) has increased: this means that heat has been transferred INTO the system from the environment (the heat came from the sun).

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Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

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$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

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Answer:

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221.55 Nm

Explanation:

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C = Drag coefficient = 1.1

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v = Speed of wind = $\frac{150}{3.6}\ m/s$

L = Length of pole

Density

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Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

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Answer:

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(b) When the angle, θ = 90 ⁰, force = 4.7 N

Explanation:

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$F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N$

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