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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.

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In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago

A phone cord is 2.28 m long. The cord has a mass of 0.2 kg. A transverse wave pulse is produced by plucking one end of the taut

Debora [2.8K]

The characteristics of the speed of the traveling waves allows to find the result for the tension in the string is:

T = 10 N

The speed of a wave on a string is given by the relationship.

v = $\sqrt{\frac{T}{\mu } }$

Where v es the velocty, t is the tension ang μ is the lineal density.

They indicate that the length of the string is L = 2.28 m and the pulse makes 4 trips in a time of t = 0.849 s, since the speed of the pulse in the string is constant, we can use the uniform motion ratio, where the distance traveled e 4 L

v = $\frac{d}{t}$

v = $\frac{4 L}{t}$

v = $\frac{4 \ 2.28 }{0.849}$

v = 10.7 m / s

Let's find the linear density of the string, which is the length of the mass divided by its mass.

μ = $\frac{m}{L}$

$\mu = \frac{0.2}{2.28}$

μ = 8.77 10⁻² kg / m

The tension is:

T = v² μ

Let's calculate

T = 10.7² 8.77 10⁻²

T = 1 0 N

In conclusion using the characteristics of the velocity of the traveling waves we can find the result for the tension in the string is:

T = 10 N

Learn more here: brainly.com/question/12545155

7 0

3 years ago

Identify the following joint

MissTica

This is a ball and socket joint.

7 0

3 years ago

What is electricity?

const2013 [10]

Explanation:

electricity is the flow of electrical power or charge

4 0

3 years ago

Read 2 more answers

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

4 years ago