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galina1969 [7]
3 years ago
9

WILL MARK BRAINLIEST JUST PLEASE HELP

Physics
2 answers:
mixer [17]3 years ago
6 0

Answer:

40N in either direction is the answer

Black_prince [1.1K]3 years ago
5 0
The Third choice 40n in either direction will be your answer. Hope this helped. Have a great day!
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52.13 dg = ___________ mg <br> someone please help me
NNADVOKAT [17]
5213 mg is the answer
8 0
3 years ago
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Calculate the amount of heat needed to raise 1.0 kg of ice at -20 degrees Celsius to steam at 120 degree Celsius
CaHeK987 [17]

Answer:

801.1 kJ

Explanation:

The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.

The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m =  mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20  °C = 4216 J

The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m =  mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100  °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m =  mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20  °C = 39920 J

The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J

+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ

4 0
3 years ago
HELP ASAP PLZ
vredina [299]

Answer:

d

Explanation:

the atomic number and the mass number must not change. so you're just looking for the particle that makes the numbers on both sides the same

7 0
3 years ago
Read 2 more answers
1. A 1000 kg equipment lift travels up a 200 m shaft in 45 seconds. Assuming the speed is constant, what is the
ANTONII [103]

Answer:

43,555 W

Explanation:

Since the speed is constant, this means that the force applied to lift the equipment is equal to the weight of the equipment. So we can write:

F=mg

where

m = 1000 kg is the mass

g=9.8 m/s^2 is the acceleration due to gravity

So,

F=(1000)(9.8)=9800 N

Then the work done in lifting the equipment is:

W=Fd

where

d = 200 m is the displacement of the equpment

Substituting,

W=(9800)(200)=1.96\cdot 10^6  J

Finally, the power used to lift the equipment is the ratio between the work done and time taken:

P=\frac{W}{t}

where

W=1.96\cdot 10^6 J

t = 45 s is the time taken

Solving,

P=\frac{1.96\cdot 10^6}{45}=43,555 W

8 0
3 years ago
A ball has mass of 140g. what is the force to accelerate the ball at 25m/s
Fittoniya [83]

We know, F = m * a

Here, m = 140 g = 0.140 Kg

a = 25 m/s2

It would be: F = 0.140 * 25 = 3.5 N

So your answer would be 3.5N


6 0
3 years ago
Read 2 more answers
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