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3 years ago

A bus accelerates from 5.75 m/s at a rate of 1.25 m/s/s for 3.50

Physics

1 answer:

marysya [2.9K]3 years ago

3 0

Answer:

10.125 meters?

Explanation:

Im taking 5.75m/s + 1.25 m/s/s (3.5) = my answer.

In those 3.5 seconds it travels 4.375.

I added that to 5.75 to get 10.125m

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4. How much time does it take for a student running at a speed of 5 m/s to cover a distance of 2,000 m?

Monica [59]

Answer:

6 Minutes 40 Seconds or 400 Seconds

Explanation:

Time to cover a distance of 5m = 1 Second

Time to cover a distance of 2000m = 2000÷5

= 400 Seconds

After converting 400 Seconds into minutes it will become 6 minutes 40 seconds.

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3 years ago

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Answer:

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4 years ago

A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a

photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

$f'=f(\frac{v-v_o}{v+v_s})$ (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

$v=\lambda f\\\\$

v: speed of sound

λ: wavelength = 0.700 m

$f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz$

Next, you replace the values of all parameters in the equation (1):

$f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz$

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

$I=\frac{P}{4\pi r^2}$

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

$I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}$

hence, the intensity is 8.12*10^⁻4 W/m^2

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3 years ago

The pressure in Arabian sea is much bigger than near the surface?<br> please don't scam

sashaice [31]

Answer:

In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing

stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.

Explanation:

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3 years ago

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A gun has a muzzle speed of 80 meters per second. What angle of elevation should be used to hit an object 180 meters away? Negle

olga55 [171]

Knowing the initial velocity and angle, the horizontal range formula is given by R= V^2sin(2teta) / g, so we can get
sin(2teta)=Rg/V^2
sin(2teta)= (180 x 9.8)/ 80^2= 0.27, sin(2teta)=0.27, 2teta=arcsin(0.27)=15.66, so teta=15.66/2

teta=7.83°

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4 years ago

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