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3 years ago

The Northern Hemisphere is warmer in spring than in winter, because in spring

Physics

2 answers:

Allisa [31]3 years ago

6 0

Answer:

Both hemispheres are warmer in their Spring than they are

in their Winter, because . . .

A). the sun climbs higher in the sky during Spring than it does during

WInter ... shooting its rays more directly at the ground ...,

and

B). the sun stays up in the sky longer in Spring than it does in

Winter, giving the ground more time to absorb its rays.

Read more on Brainly.com - brainly.com/question/788447#readmore

Explanation:

Amanda [17]3 years ago

4 0

Both hemispheres are warmer in their Spring than they are
in their Winter, because . . .

A). the sun climbs higher in the sky during Spring than it does during
WInter ... shooting its rays more directly at the ground ...,

and

B). the sun stays up in the sky longer in Spring than it does in
Winter, giving the ground more time to absorb its rays.

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Which landform represents the boundary between the land and an ocean or a lake?

Likurg_2 [28]

Coastline or seashore

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3 years ago

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A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran

marishachu [46]

<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train = 4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

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3 years ago

Define electric current and drift velocity.

Blizzard [7]

Answer:

Current- the flow of free charges, such as electrons and ions

Drift velocity- the average speed at which these charges move

3 0

2 years ago

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A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

2 years ago

A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he

RideAnS [48]

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2} (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

4 0

3 years ago