The ozone layer is like the glass. the light from the sun enters the atmosphere and when the heat bounces back it is hard to escape the ozone layer therefore the earth's temperature rises

5 0

3 years ago

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What is the momentum of a 5 kg object that has a velocity of 1. 2 m/s? 3. 8 kg â€˘ m/s 4. 2 kg â€˘ m/s 6. 0 kg â€˘ m/s 6. 2 kg â

nevsk [136]

The momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

<h3> MOMENTUM:</h3>

Momentum of a substance is the product of its mass and velocity. That is;

Momentum (p) = mass (m) × velocity (v)

According to this question, an object has a mass of 5kg and velocity of 1.2m/s. The momentum is calculated thus:

Momentum = 5kg × 1.2m/s

Momentum = 6kgm/s.

Therefore, the momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

Learn more about momentum at: brainly.com/question/250648?referrer=searchResults

7 0

2 years ago

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The density of a substance equals its mass divided by its volume. Talia listed the density of some common materials at 20 °C.

Zolol [24]

Answer:

mercury

Explanation:

6 0

3 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

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

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

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

6 0

3 years ago