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3 years ago

A truck decelerates at a rate of 2 m/s2. If it starts with a velocity of 15 m/s, how long will it take to stop?

1 answer:

7 0

Use the formula Vf = Vo + at

Stop => Vf = 0

=> 0 = Vo + at => Vo = -at => t =- Vo / a

Vo = 15 m/s

a = - 2 m/s^2

=> t = - 15 m/s / ( - 2 m/s^2) = 7.5 s

And you can find the distance travled by using the equation: d = Vo*t + at^2 / 2

=> d = 15 m/s * 7.5 s - 2m/s^2 * (7.5s)^2 / 2 = 56.25 m

Answer: 7.5s (and 56.25m)

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2 years ago

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61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

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3 years ago

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The frequency of the electromagnetic wave is 9.55 × 1014 Hz and it is classified as ultraviolet.

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