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2 years ago

13

4.2.5 quiz waves and technology

1 answer:

kolezko [41]2 years ago

6 0

Answer:

What’s the Question

Explanation:

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A plane is flying horizontally with speed 292 m/s at a height 3880 m above the ground, when a package is dropped from the plane.

kap26 [50]

The plane is straight above the package when it hits the ground.
Find in what way long a free fall from a height of 4240 m takes: s = 1/2 gt^2
t^2 = 2s/g
t^2 = 2* 3880/9.8 = 791.8367
t = √791.8367
t = 28.1396 seconds
Now find the horizontal displacement with v = 292 m/s in 28.1396 s:
s = v*t
s = 292 * 28.1396 = 8216.76 m

4 0

3 years ago

How many significant digits are in the measurement 50.003010

Anastasy [175]

Hello There!

It is 6 sf

Hope This Helps You!
Good Luck :)

- Hannah ❤

6 0

3 years ago

A spring with spring constant 13.1 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Alenkinab [10]

Answer:

Explanation:

spring constant, K = 13.1 N/m

22 oscillations in 20 seconds

time taken to complete one oscillation is called time period.

T = 20 / 22 second = 0.909 seconds

(a) let m be the mass.

The formula for the time period is

$T = 2\pi \sqrt{\frac{m}{K}}$

$m = \frac{T^{2}K}{4\pi ^{2}}$

$m = \frac{0.909^{2}\times 13.1}{4\pi ^{2}}$

m = 0.275 kg

(b) maximum speed, v = ω A = 2π A / T

v = ( 2 x 31.4 x 0.1) / 0.909

v = 0.691 m/s

6 0

3 years ago

An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what

muminat

Answer:

The density is $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

The weight in air is $W_a = 40 \ N$

The weight in water is $W_w = 20 \ N$

The weight in a unknown liquid is $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

$W_w = W_a -m _w g$

Where $m_w$ is he mass of the water displaced

substituting value

$m_w g = 40 -20$

$m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown is mathematically represented as

$W_u = W_a -m _u g$

Where $m_u$ is he mass of the unknown liquid displaced

substituting value

$m_u g = 40 -30$

$m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

$\frac{m_ug}{m_wg} = \frac{10}{20}$

$\frac{m_u}{m_w} = \frac{1}{2}$

=> $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

$\rho _u = 0.5 \rho_w$

This because

$density = \frac{mass}{volume}$

So if volume is constant

mass = constant * density

Where $\rho_u$ is the density of the liquid

and $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

So

$\rho_u = 1000*0.5$

$\rho_u =500 kg /m^3$

7 0

3 years ago

Select the correct text in the passage.

serious [3.7K]

Answer:

I have no clue

Explanation:

i just have no clue

4 0

3 years ago

Read 2 more answers