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zvonat [6]
2 years ago
13

A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C. How much water freezes onto the ice? The specific heat of

ice is 0.5 cal/g · ◦ C and its heat of fusion of is 80 cal/g. Answer in units of g.
Physics
1 answer:
vladimir2022 [97]2 years ago
4 0

For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice?  is mathematically given as

x = 9.93 g

<h3>What is the amount of water that freezes onto the ice?</h3>

Where

Energy received = energy given out

Generally, the amount of water is mathematically given as

(53)(0.5)(30) = (80)(x)

Therefore

x = (49)(0.5)(16)/(80)

x = 9.93 g

In conclusion, the mass of water

x = 9.93 g

Read more about  mass

brainly.com/question/15959704

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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio
Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

If we know that m = 50\,kg and g = 9.807\,\frac{m}{s^{2}}, the weight of the sled is:

W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

F_{min,k} = 49.035\,N

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0
3 years ago
If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w
Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

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3 years ago
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