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djyliett [7]
3 years ago
12

How do meteorites differ from meteors and meteoroids?

Physics
2 answers:
Evgesh-ka [11]3 years ago
8 0

Answer:

When meteoroids enter Earth's atmosphere (or that of another planet, like Mars) at high speed and burn up, the fireballs or “shooting stars” are called meteors. When a meteoroid survives a trip through the atmosphere and hits the ground, it's called a meteorite.

Explanation:

Ira Lisetskai [31]3 years ago
6 0

Answer: Meteoroid: Small particle from a comet or asteroid orbiting the Sun. Meteor: The light phenomena which results from a meteoroid entering the Earth's atmosphere and vaporizing; this is basically a shooting star. Meteorite: A meteoroid that survives through the Earth's atmosphere and lands on the Earth's surface.

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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
The index of refraction of light varies from color to color. True or False?
lys-0071 [83]
The index of refraction of light varies from color to color. TRUE.
8 0
3 years ago
Sally goes on a hike. The distance she walks compared to the amount of time she walks is shown on the graph. Which statement is
Neko [114]
According to the task there should be the graph that supports Sally's hike, but after looking on the options it seems that Sally doesn't walks at a constant rate and there is the negative option that coincides with my thoughts. So, I bet the false statement is the third option represented in the scale above.
5 0
3 years ago
You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract
stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

Learn more about Velocity here, brainly.com/question/18084516

#SPJ4

3 0
2 years ago
A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

                    or,  y =  3.56 cm.

3 0
3 years ago
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