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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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The other person who answered this is wrong btw
5 0
3 years ago
A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.
Inessa [10]

Hi there!

We can use Newton's Second Law:
\Sigma F = ma

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}

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2 years ago
A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its
larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

F_n_e_t=ma

Plug in what you know, and solve:

18=m(12)\\m=1.5kg

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C.) 1.5 kg

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2 years ago
A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

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3 years ago
What is the meaning of smooth?
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