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Dima020 [189]
4 years ago
6

When all parts of a circuit are composed of conducting materials, the circuit is said to be

Physics
1 answer:
Korolek [52]4 years ago
7 0
Closed is the correct answer :)
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A particle moving on a plane curve what is the degree of freedom
kirill115 [55]

A particle confined to move along a curved path has only one degree of freedom. inclined plane are some examples of constrained motion. Every condition of constraint reduces the number of degree of freedom by one.

I hope this helps!

7 0
3 years ago
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
T/F: Stars die.<br> True<br> False
Yuki888 [10]

Answer:

The answer is true

Explanation:

could you brainliest again they said I plagarized when I didn't

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What is also known as watered carbons
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The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.
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What are the equations of angular acceleration, angular velocity and theta?.
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Θ is the angular displacement = ωt
ω is the angular velocity = θ/t
α is the angular acceleration = ω/t
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3 years ago
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