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2 years ago

15

How much does a 60 kg table weigh

1 answer:

soldier1979 [14.2K]2 years ago

5 0

Answer:

588 N

Explanation:

weigh = Mg

I hope you got that. Thanks and upovote that

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ns:Select the correct answer. In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hi

Len [333]

You would need to use the equation a= (v-u)÷t
You need to substitute in the correct numbers.
a= (10-20)÷1
Your answer is -10m/s^2

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2 years ago

Calculer l intensité du courant qui le traverse

MissTica

Answer:

Le calcul du courant se fait avec deux éléments : la tension et la valeur de la résistance. Courant (A) = tension (V) / résistance (Ohm) ce qui donne la formule I = U/R.

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2 years ago

How do you break apart a

Tcecarenko [31]

Answer:

B

Explanation:

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1 year ago

Read 2 more answers

Can an object be in motion and not in motion at the same time? Explain.

kolezko [41]

It could rotate while not advancing distance

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2 years ago

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

2 years ago