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Andreas93 [3]
3 years ago
8

A person run 21.0 km west then runs 10.0 km east what is the persons distance

Physics
2 answers:
VashaNatasha [74]3 years ago
6 0

Answer:

31.0 km

Explanation:

21.0+10.0=31.0 km

Andrej [43]3 years ago
5 0

Answer:

Distance from starting point = 11 km, Distance ran altogether = 31 km

Explanation:

21 - 10 = 11

21 + 10 = 31

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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
2 years ago
In a lab,two pieces of cellphane tape are placed on a lab table. They are quickly ripped off the lab table. Which would you pred
galben [10]

Answer:

they would repel each other

Explanation:

they have gained the same charge

4 0
3 years ago
Can waves move or transport matter? Why or why not?
OLga [1]
They can.
Surfers use waves to move themselves around.
6 0
2 years ago
Identify the Following as physical properties or chemical properties.
sveticcg [70]


A.) is chemical, B.) is physical, C.) is physical, D.) is chemical, E.) is physical, F.) is physical, G.) is physical, and H.) is chemical.

5 0
2 years ago
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A 720 g softball is traveling at 15.0 m/s when caught. If the force of the glove on the ball is 520 N, what is the time it takes
Sholpan [36]

Answer:

The time it takes the ball to stop is 0.021 s.

Explanation:

Given;

mass of the softball, m = 720 g = 0.72 kg

velocity of the ball, v = 15.0 m/s

applied force, F = 520 N

Apply Newton's second law of motion, to determine the time it takes the ball to stop;

F = ma = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{0.72 \ \times \ 15}{520} \\\\t = 0.021 \ s \\

Therefore, the time it takes the ball to stop is 0.021 s.

5 0
3 years ago
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