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Ugo [173]
3 years ago
5

A diamond with a mass of 45 g hangs motionless from a chain. what is the upward force of the chain on the diamond?

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0
The upward force of the chain on the diamond would be the tension in the chain, and this tension would have to support the weight of the 45g that hangs from the chain.

mass = 45 g = 45/1000 kg = 0.045kg

Weight = mg = 0.045 * 10 ≈ 0.45N,            g ≈ 10 m/s²

<span>So the upward force is ≈ </span><span>0.45N. </span>
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How does the circuit change when the wire is added? a closed circuit occurs and makes all bulbs turn off. an open circuit occurs
Vitek1552 [10]

The circuit change when a wire is added is, an open circuit occurs and makes all bulbs turn off.

<h3>What is a closed circuit?</h3>

A closed circuit is a type of circuit connection in which the wire connection is complete and current flow occurs, turning the light bulbs on in the process.

<h3>What is an open circuit?</h3>

An open circuit is a type of circuit connection in which the wire connection is incomplete and current cannot flow, turning off the light bulbs.

Thus, the circuit change when the wire is added is, an open circuit occurs and makes all bulbs turn off.

Learn more about open circuit here: brainly.com/question/20351910

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3 0
2 years ago
Two substances, M and N, have specific heats c and 2c. if heats Q and 4Q are supɔlied to Mand N, respectively, their changes in
Ierofanga [76]

Answer:

If the mass of B is m and the temperature change is the same, the mass of B will be 2m.

Explanation:

Q = mcT

T = mc/Q

M = 4Q/2cT........... (1)

T = Q/mc

Plug this in equation 1.

M = 4Q/(2c × Q/mc)  = 4Q ÷ 2Q/m  = 4Q × m/2Q = 2m

6 0
2 years ago
The image formed by a plane mirror is _____.
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4 0
3 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
3 years ago
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The north vectors add up as so the south vectors.  Then subtract the two.  For north its 4 + 5 = 9.  South is 2 + 5 = 7.   Then 9-7 = 2km North (D)
7 0
3 years ago
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