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3 years ago

Help help help help help

Physics

2 answers:

Aleksandr [31]3 years ago

6 0

Explanation:

P= rho×g × h (g=10 m/s²) ,

(rho=1000 kg/m³)

P= 1000×10×5= 50,000 N/m² (C)

Anvisha [2.4K]3 years ago

5 0

Answer:

C.50,000 N/m²

Explanation:

P=rho×g×h

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The balloon's behavior results from a change at the electron level. What do you think might have happened while rubbing the ball

Mariulka [41]

Answer:

Explanation:

opposite charges attract each other, while rubbing the objects together they developed opposite charges which caused them to drift together

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4 years ago

A hockey puck sliding at 60.0m/s slows uniformly to 20.0m/s while travelling 800.0m.

julia-pushkina [17]

A) -3.75 meters/second
A=(20^2-80^2)/(2x800)
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3 years ago

Pls help me with this one

lora16 [44]

Answer :

The answer is clearly C

Explanation:

Because the only way currents move are to the side

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3 years ago

A box took 5 minutes to move you did 2150 j of work in the process. how much power was required to move the box

AURORKA [14]

Power = Work/Time

Convert 5 min to seconds and divide 2150J by that number.

4 0

3 years ago

Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl

tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

$v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s$

For Train 2 Velocity of the Source,

$v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s$

Frequency of Source,

$f_{s}=500\ Hz$

To Find:

Frequency of Observer,

$f_{o}=?$ (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source $v_{s}$ and observer $v_{o}$ , the original frequency of the sound waves $f_{s}$ and the observed frequency of the sound waves $f_{o}$ ,

The Equation is

$f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})$

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

$f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz$

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

7 0

3 years ago