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IgorLugansk [536]

3 years ago

11

If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder wil change from 70

kilopascals to what?

1 answer:

ira [324]3 years ago

4 0

The pressure should increase to 140kPa

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Physics question, answer completely with work

Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

$I=F\Delta t = \Delta p$

Where

I is the impulse

F is the force

$Delta t$ is the time interval during which the force is applied

$\Delta p$ is the change in momentum

In this problem,

$\Delta t = 3.0 s$ is the time interval

$I=6.0 N\cdot s$ (east) is the impulse

Therefore, the magnitude of the force is

$F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N$

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0

3 years ago

1) A force of 20 Newton acts on a bar having a cross sectional area of 0.8m^2 and length 10cm.calculate the stress developed in

Elanso [62]

Answer:25N/M^2

Explanation:

Force=20N Area=0.8M^2

Stress=force/area

Stress=20/0.8

Stress=25N/M^2

4 0

3 years ago

A jet - powered car called the spirit of America required 9600meters to stop from its highest speed . If the car decelerated at

Mazyrski [523]

$v^{2}$ = $u^{2} + 2ar$
0 = u^2 + 2*(-2)*9600
u^2 = 38400
u = 195.96 m/s

6 0

3 years ago

True or False: An example of an Isometric movement is a plank. please help 2 mins left!<br> PE

Anon25 [30]

Answer:

Yes

Explanation:

The plank (also called a front hold, hover, or abdominal bridge) is an isometric core strength exercise that involves maintaining a position similar to a push-up for the maximum possible time

5 0

3 years ago

Read 2 more answers

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

$a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m
\\d=150rev$

$d= 155 rev = 155(2\pi ) = 310\pi rad$

$a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi ) = 4.0\pi \frac{rev}{s^{2} }$

$d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449$

$w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}$

Now the angular velocity is the blade speed so:

$V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}$

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

$t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s$

Now is the same time the piece travel horizontally

$L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m$

blade piece travels HORIZONTALLY = (24.5)(0.397) = 9.73 m ANS

6 0

3 years ago