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Helga [31]
3 years ago
6

Raindrops are falling straight down at 11 m/s when suddenly the wind starts blowing horizontally at a brisk 5.0 m/s. From your p

oint of view, the rain is now coming down at an angle. What is the angle, relative to the vertical?
Physics
1 answer:
Bumek [7]3 years ago
8 0

Answer: 24.4 degrees to the vertical

Explanation:

Vertical component of raindrop speed = 11m/s

Horizontal component of wind = 5m/s

In this case, all we have to do is to use trigonometric ratios of angles to sides as in a triangle

Doing this, we see that

tan (theta) = 5/11

(Where theta is the angle made with the vertical by the rain after impact)

Tan being opposite/adjacent

Arc sin (5/11) gives us 24.44 degrees to the vertical

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Mekhanik [1.2K]

Answer:

<h3>If we add impurity into a matter then elasticity</h3><h2>May change</h2>

Explanation:

When we add impurity into a matter then elasticity of the substance increases.

3 0
3 years ago
12. A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of t
Kitty [74]

Answer:

Explanation:

F = GmM/d²

As gravity force is proportional to the inverse of the square of the distance,

doubling the distance will reduce the weight to a forth.

F' = GmM/(2d)²

F' = ¼GmM/d²

F' = ¼F = ¼(4000)

F' = 1000 N

6 0
3 years ago
A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh
eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}

P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}

P^2 = \frac{P^2V^2}{c^2}+m^2V^2

P^2 = V^2 (\frac{P^2}{c^2}+m^2)

V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}

V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}

V = 2.81784*10^8m/s

Therefore the height with respect the observer is

l = l_0*\sqrt{1-\frac{V^2}{c^2}}

l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}

l = 0.56m

Therefore the height which the observerd measure for her is 0.56m

8 0
3 years ago
An Olympic-class sprinter starts a race with an acceleration of 5.10 m/s2. What is her speed 2.40 s later?
ivolga24 [154]

Answer:

12.24 m/s

Explanation:

Speed: This can be defined as the rate of change of distance with time. The S.I unit of speed is m/s.

Using the formula,

a = v/t................ Equation 1

Where a = acceleration of the sprinter, v = speed of the sprinter, t = time.

making v the subject of the equation,

v = at ................. Equation 2

Given: a = 5.1 m/s², t = 2.4 s.

Substitute into equation 2

v = 5.1(2.4)

v = 12.24 m/s.

Hence, the speed of the sprinter = 12.24 m/s

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3 years ago
New alleles arising from mutations in a population will​
ivanzaharov [21]
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