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MariettaO [177]
3 years ago
5

To maintain the same amount of torque due to a mass on a balance as the mass is increased, how should the position of the mass c

hange
Physics
1 answer:
olasank [31]3 years ago
8 0

Answer:

the mass should be bring closer to the point about which we are finding torque

Explanation:

τ = Σr × F = rmg

where m is the mass, g is acceleration due to gravity, and r is the distance

Torque is directly proportional to -

1.mass, m , of object

2. distance, r, of the mass from the point about which we are finding the torque.

So if we increase or decrease them then the torque will also increase or decrease.

So if we increase the mass the torque will increase but since we have to maintain same torque therefore we have to decrease the distance of mass from the point about which we are finding torque.

Therefore the mass should be bring closer to the point about which we are finding torque.

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A three-way 120 V lamp bulb that contains two filaments is rated for 100-200-300 W. One filament burns out. Afterward, the bulb
ololo11 [35]

Answer:

The larger resistance value is 144 ohms and the smaller resistance value is 48 ohms.

Explanation:

Given that Resistance, R, is inversely proportional to Power, the largest value of power will yield the smallest value of resistance and vice versa.

The largest value of resistance will be:

P = V²/R

R = V²/P

R = 120²/100

R = 144 ohms

The smallest value of resistance will be:

R = 120²/300

R = 48 ohms

8 0
4 years ago
Please help quick please its a test
navik [9.2K]

Answer:

Im pretty sure its y!

Explanation:

6 0
3 years ago
A pupil adds 37g of ice at 0°C to 100g of water at 30°C. The final temperature of the water and melted ice is 0°C. no heat is lo
mojhsa [17]

Answer:

The specific latent heat of ice is approximately 341 J/g

The correct option is;

b) 341 J/g

Explanation:

The given parameters are;

The mass of ice the pupil adds to the water, m₁ = 37 g

The initial temperature of the ice, T₁ = 0°C

The mass of the water to which the ice is added, m₂ = 100 g

The initial temperature of the water, T₂₁ = 30°C

The final temperature of the water and the melted ice, T₂₂ = 0°C

The specific heat capacity of the water, c₂ = 4.2 J/(g·°C)

By the principle of conservation of energy, we have;

The heat gained by the ice = The heat lost by the water = ΔQ₂

Given that the ice is only melted with no change in temperature, we have;

The heat gained by the ice = The latent heat needed for melting the ice

ΔQ₂ = m₂ × c₂ × (T₂₂ - T₂₁) = 100 × 4.2 × (0 - 30) = -12,600 J

The heat gained by the ice = m₁ × L_f

Where;

L_f represents the specific latent heat of fusion of ice;

We have;

12,600 = 37 × L_f

L_f = 12,600/37 = 340.54 J/g ≈ 341 J/g

The specific latent heat of fusion of ice = L_f ≈ 341 J/g.

5 0
3 years ago
A tertiary source is the most desirable source for a researcher to use. True or false
Kaylis [27]

The statement ‘A tertiary source is the most desirable source for a researcher to use’ is false. It should be that your source is primary because it is more reliable and is related to the research you have been working on.

6 0
3 years ago
A race car is one lap behind the lead race car when the lead car has 44 laps to go in a race. If the speed of the lead car is 55
masha68 [24]

Answer:

Speed of second car will be 57.17 m/sec

Explanation:

We have given lead car travels = 44 laps

1 laps = 1.34 km = 1340 m

So total distance = 1340×44 = 58960 m

Speed of lead car = 55.9 m/sec

We know that time=\frac{distance}{speed}=\frac{58960}{55.9}=1054.74sec

As the second car is 1 lap behind so distance traveled by second car = 45×1340 = 60300 m

So speed of second car will be speed=\frac{distance}{time }=\frac{60300}{1054.74}=57.17m/sec

3 0
3 years ago
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