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3 years ago

HELP WITH THIS QUESTION PLEASE! What Factors Affect Climate?

2 answers:

7 0

The climate of any particular place is influenced by a host of interacting factors. These include latitude, elevation, nearby water, ocean currents, topography, vegetation, and prevailing winds. The global climate system and any changes that occur within it also influence the local climate.

harkovskaia [24]3 years ago

5 0

Latitude, elevation, ocean currents, topography, and prevailing winds. There's probably a few others but these are the most important.

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I need help with...

Rom4ik [11]

The car will gain new momentum if it's velocity is doubled or tripled.

6 0

4 years ago

A proton moves with a speed of 1.17 105 m/s through Earth's magnetic field, which has a value of 50.0 µT at a particular locatio

vlabodo [156]

a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.

b) Given the equation for Magnetic Force

$F= qVB$

Replacing

$F= (1.16*10^{-19})(1.17*10^5)(50*10^{-6})$

$F=9.36*10^{-19}$

c) Given the second Newton's Law by

$F_g = 1.67*10^{-27}*9.81$

$F_g = 1.64*10^{-26}$

Given the electric force by,

$F_e = 1.6*10^{-19}*1.5*10^2$

$F_e = 2.4*10^{-17}N$

$F=9.36*10^{-19}N$

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

Calculate the acceleration of a train travelling from rest to 24 m/s in 12 seconds.

Vilka [71]

Answer: A 2m/s^2

Steps: Formula for acceleration. (Velocity Final - Initial Velocity) / Time

(24 - 0) / 12 = 2

3 0

3 years ago

What measurements or observations tells you that a car is accelerating

lutik1710 [3]

You need to observe the car at two different times.

-- The first time:
You write down the car's speed, and the direction it's pointing.

-- The second time:
You write down the car's speed and the direction it's pointing, again.

You take the data back to your lab to analyze it.

-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.

-- You compare the first and second direction. If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.

(Remember, "acceleration" doesn't mean "speeding up".
It means any change in speed or direction of motion.)

8 0

4 years ago

Read 2 more answers