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andreyandreev [35.5K]

3 years ago

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A 3.0-kg block moves up a 40o incline with constant speed under the action of a 26-N force acting up and parallel to the incline

. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

1 answer:

CaHeK987 [17]3 years ago

3 0

Answer:

it is very hard question for me sorry i cant solve it

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A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,

masha68 [24]

Answer:

a) The car was moving at a speed of $29.167\ m.s^{-1}$

b) The negative sign of $v_o$ denotes that the observer is coming towards the police car which is the source of the sound.

c) $f_o=1283.33\ Hz$

Explanation:

Given:

original frequency of the source, $f=1200\ Hz$

speed of the source, $v_s=0\ m.s^{-1}$

velocity of the obstacle car be, $v_o$

speed of sound, $s=350\ m.s^{-1}$

observed frequency, $f_o=1100\ Hz$

<u>Using the equation from the Doppler's effect:</u>

$\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}$

$\frac{1100}{1200} =\frac{(350+v_o)}{350-0}$

$v_o=-29.167\ m.s^{-1}$

a)

The car was moving at a speed of $29.167\ m.s^{-1}$

b)

The negative sign of $v_o$ denotes that the observer is coming towards the police car which is the source of the sound.

c)

Now when, $v_s=50\ m.s^{-1}$

Then, $f_o=?$

Using the Doppler's eq.:

$\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}$

$f_o=1283.33\ Hz$

3 0

3 years ago

7. Navy use SONAR method to locate the position of submarine under water. The waves used in SONAR are A transverse waves. B elec

pshichka [43]

Answer:

A. transverse waves

vote me brainliest?? 5/5

thanks:)

8 0

3 years ago

Read 2 more answers

Tell the value of the underlined digit 843,208,732,833 eight is underlined

devlian [24]

The value of the underlined 8 is, hundred billion's. Hope this helped!

4 0

3 years ago

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

The pressure of air pushing down on a house roof is extremely large. Why is the roof not crushed?

Degger [83]

There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.

7 0

3 years ago