All categories

4 years ago

Why are average temperatures higher at the equator than near the poles?

2 answers:

8 0

The poles are on either side of the sun so therefore the equator is constantly the closest to the sun and gets the most heat

elena55 [62]4 years ago

3 0

Bc the equator is constantly facing the sun while the earth rotates

You might be interested in

Which of the following is an example of speed? Question 8 options: A truck driving 15 miles per hour south. A cat running 5 mete

xxTIMURxx [149]

The answer would be <u>a cat running at 5 meters per minute</u>.

Hope this helps :)

7 0

4 years ago

Read 2 more answers

The color in visible light have different

alina1380 [7]

<span>The different colors of visible light have different wavelengths. </span>

5 0

3 years ago

Look at Concept Simulation 5.2 to review the concepts involved in this question. Two cars are identical, except for the type of

dimulka [17.4K]

Answer:

Tread design on car B would yield a larger coefficient of static friction between the tires and the road

Explanation:

The car model using the coefficient of static friction doesn't work well with tires. A higher coefficient of static friction would require more force to cause a loss of attraction.

The static frictional force helps to keep the unbanked horizontal turn. This means that the frictional force is the centripetal force.

The tread design of car B ensures that the centripetal force is enough to negotiate the turn. On the other hand, the tread design of car A does not provide the necessary centripetal force, hence car A is unable to negotiate the turn.

Therefore, tread design on car B would yield a larger coefficient of static friction between the tires and the road.

7 0

4 years ago

A 900 kg car runs into a 70 kg deer at 28 m/s. If the car transfers all of its momentum to the deer, how fast does the deer go f

Alex

Answer:

Mc = 900 Kg

Uc = 28 ms^-1

Md = 70 Kg

Ud = 0

We want Vd

Vc = Vd

This situation is elastic momentum

$m _{c}u _{c}+ m _{d}u _{d} = m _{c}v _{c} + m _ dv _{d} \\ (900 \times 28) + (70 \times 0) = (900 \times v _{d} ) + (70 \times v _{d}) \\ 25200 = (900 + 70)v _{d} \\ 25200 = 970v _{d} \\ v _{d} = \frac{25200}{970} \\ v _{d} = 26 \: m {s}^{ - 1}$

4 0

3 years ago

With what tension must a rope with length 2.90 m and mass 0.125 kg be stretched for transverse waves of frequency 42.0 Hz to hav

Vesnalui [34]

Answer:

41.64 N

Explanation:

Applying,

v = √(T/m')................ Equation 1

Where v = velocity of the wave, T = Tension of the rope, m' = mass per unit length of the rope.

make T the subject of the equation,

T = v²m'................. Equation 2

But,

v = λf............... Equation 3

Where λ = wavelength, f = frequency

And

m' = m/L........... Equation 4

Where m = mass of the rope, L = length of the rope

Substitute equation 3 and equation 4 into equation 2

T = (λf)²(m/L).............. Equation 5

From the question,

Given: λ = 0.740 m, f = 42 Hz, m = 0.125 kg, L = 2.9 m

Substitute these values into equation 5

T = (42×0.74)²(0.125/2.9)

T = 41.64 N

6 0

3 years ago