A DC voltmeter may be used directly to measure a) frequency b)polarity c) power factor d) power

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

What momentum of a 50kilogram ice skater gliding across the ice at a speed of 5m/s?

Ivenika [448]

Momentum (P) = Mass (kg) * Velocity (m/s)

P = M * V
P = 50 * 5
P = 250

So momentum is 250 kgm/s

5 0

3 years ago

Read 2 more answers

An electric motor rotates 60 times per second if the alternating current source is 60 Hz. How many times will an electric motor

valentina_108 [34]

Answer:

180,000

Explanation:

Frequency is a quantity that is measured in Hertz [Hz] and it represents the number of rotations per second.

A motor with a frequency of 50 Hz will rotate 50 times per second.

Since we don't want to know how many times it rotates per second, but per hour. The first step is to find how many seconds there are in an hour and then multiply that amount by 50.

Seconds in an hour:

there are 60 seconds per minute, and 60 minutes per hour, thus there are

60*60 = <u>3,600 seconds in an hour</u>

We know that the motor will rotate 50 times per second so to find the number of rotations in 1 hour = 3,600 seconds we multiply:

50*3,600 = 180,000 rotations

8 0

3 years ago

A container is ﬁlled with water to a depth of 26.2 cm. On top of the water ﬂoats a 16.1 cm thick layer of oil with a density of

Solnce55 [7]

Answer:

1.34 x 10^3 Pa

Explanation:

density of oil = 0.85 x 10^3 kg/m^3

g = 9.81 m/s^2

height of oil column = 16.1 cm = 0.161 m

Pressure on the surface of water = height of oil column x density of oil x g

= 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa

Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.

4 0

3 years ago

What force is necessary to accelerate a 5.0 kg mass from rest to a final velocity of 10.0 m/s in 5.0 s?

vesna_86 [32]

Answer:

10 N

Explanation:

F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N

4 0

3 years ago