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3 years ago

1. A gas is kept at a pressure of 22Kpa and a temperature of 287K. What will

Physics

1 answer:

Ann [662]3 years ago

6 0

Answer:

New pressure (P2) = 22.92 Kpa (Approx)

Explanation:

Given:

Old pressure (P1) = 22 Kpa

Temperature (T1) = 287 K

Temperature (T2) = 299 K

Find:

New pressure (P2)

Computation:

P1 / T1 = P2 / T2

22 / 287 = P2 / 299

New pressure (P2) = 22.92 Kpa (Approx)

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A train moving at a constant speed of 50.0 km/h moves east for 45.0 min, then in a direction 45.0° east of due north for 10.0 mi

Yuki888 [10]

Answer:

3.47km/h with a direction of 67.8 degrees north of west.

Explanation:

First we need to calculate the displacement on the X axis, so:

$d_{x1}=50.0km/h*45min(\frac{1h}{60min})=37.5km\\d_{x2}=50.0km/h*10min(\frac{1h}{60min})*cos(45^o)=5.90km\\d_{x1}=-50.0km/h*55.0min(\frac{1h}{60min})=45.8km\\D_x=37.5+5.90-45.8=-2.4km$

then on the Y axis:

$D_y=50.0km/h*10min(\frac{1h}{60min})*sin(45^o)=5.90km$

The magnitud of the displacement is given by:

$D=\sqrt{D_x^2+D_y^2} \\D=6.37km$

and the angle:

$\alpha =arctg(\frac{5.90}{2.41})=67.8^o$

that is 67.8 degrees north of west.

$v=\frac{D}{t}\\v=\frac{6.37km}{110min*\frac{1h}{60min}}\\v=3.47km/h$

8 0

3 years ago

Mr. Stephenson drives his car from Dallas to Town X, a distance of 437 km. The trip takes 7.27 hours. What is the average speed

klasskru [66]

Average speed = (distance covered) / (time to cover the distance).

Distance covered = 437 km = 437,000 meters

Time to cover the distance = (7.27 hrs) x (3600 sec/hr) = 26,172 seconds

Average speed = (437,000 meters) / (26,172 seconds)

Average speed = 16.7 m/s

8 0

3 years ago

A student wants to determine the effect of mass on kinetic energy. She will drop two balls of the same size into a pool of water

Anastasy [175]

Answer:

She should drop two balls with different masses from the same height

Explanation:

4 0

2 years ago

A Cathode of initial mass 10.00g weigh to 10.05g

Leto [7]

Answer:

i'm sorry i'm not a physics student

6 0

3 years ago

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a

Nina [5.8K]

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

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

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'\dfrac{v}{v+v_s}\\\Rightarrow 1.01f'=f'\dfrac{v}{v+v_s}\\\Rightarrow 1.01=\dfrac{343}{343+v_s}\\\Rightarrow v_s=\dfrac{343}{1.01}-343\\\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'\dfrac{v}{v+v_s}\\\Rightarrow 0.99f'=f'\dfrac{v}{v+v_s}\\\Rightarrow 0.99=\dfrac{343}{343+v_s}\\\Rightarrow v_s=\dfrac{343}{0.99}-343\\\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

3 0

3 years ago