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3 years ago

Identify where the sun's rays strike earth most directly and least directly

1 answer:

6 0

-- Depending on the time of the year, the sun's rays strike Earth
most directly somewhere between the Tropic of Cancer and the
Tropic of Capricorn.

That's a belt around the Earth's "middle" called the "Tropic Zone".
The equator is in the middle of it, the Tropic of Cancer is 23.5° North
of the equator, and the Tropic of Capricorn is 23.5° degrees South of it.

The sun's rays can never be totally direct, straight down onto the Earth's
surface, anywhere outside this belt.

-- The sun's rays strike Earth least directly wherever, and whenever,
the sun is setting.

You might be interested in

How many zebras automatically survive the first interaction with the lions in Generation 1?

lesya692 [45]

I think the answer is c. but I think it depends on how many zebras you have

5 0

4 years ago

Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha

tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

R_cm = 1 / M ∑ x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

M = m_earth + m_moon

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

M = 5.98 10²⁴ + 7.35 10²²

M = 6.0535 10₂⁴24 kg

we substitute

R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

R_cm = 4.66 10⁶ m

4 0

3 years ago

Which of these is an example of a mechanical wave?

Oduvanchick [21]

i think the answer is B police siren.

8 0

3 years ago

Read 2 more answers

car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0

3 years ago

If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in

Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0

3 years ago