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timurjin [86]
3 years ago
14

How does the motion of the molecules change when you increase the temperature

Physics
1 answer:
borishaifa [10]3 years ago
6 0

Answer:

They speed up as temperature increases.

Explanation:

As temperatures increases within a molecule the particles will speed up.

You might be interested in
What is 3.75 x 10^-7?
pogonyaev

Answer:

Explanation:

3.75 * 10^-7

=3.75 * 1/10^7

=3.75/10000000

=3/800000000

any base which has it's power negative do it's reciprocal then the power will be positive.

8 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
A crime suspect fled the scene when police officers entered the building. How far could he run in 10 minutes if he can run 5 mil
emmainna [20.7K]
(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>

(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>
3 0
3 years ago
An airplane flies 20km in a direction 60 degrees north of east, then 30 km straight east, then 10km straight north. How far and
Fed [463]

Answer:

48.4 km, 34.3° north of east

Explanation:

Let's say east is the +x direction and north is the +y direction.

Adding up the x components of the vectors:

x = 20 cos 60 + 30 + 0

x = 40 km

Adding up the y components of the vectors:

y = 20 sin 60 + 0 + 10

y = 27.3 km

The magnitude of the displacement is:

d = √(x² + y²)

d = 48.4 km

The direction is:

θ = atan(y/x)

θ = 34.3° north of east

6 0
3 years ago
Pls help
xenn [34]

Answer:

A. It must be zero

Explanation:

A spacecraft leaves the solar system at a velocity of 1,500 m/s. The net force on this spacecraft is zero. What can we say about the spacecraft's acceleration?

According to Newton's second law

Force = Mass × acceleration

If the net force is zero

0 = mass × acceleration

0 = ma

a = 0/m

a = 0m/s²

this shows that the acceleration will be zero If the net force is zero

4 0
2 years ago
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