On reducing the volume of gas at constant temperature the pressure of gas increases why

1 answer:

8 0

Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If volume increases, then pressure decreases and vice versa, when temperature is held constant.

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A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and b

Ksenya-84 [330]

We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
F = m * a

a is acceleration. To find it we use formula:
a = Δv / Δt
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N

To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N

We can see that weight is much smaller than the applied force so it's influence in negligible.

3 0

3 years ago

Explain why ventilation is very important if there is risk of exposure to random gas in your home school

bonufazy [111]

Ventilation is very important because it helps remove the gas form people’s homes and schools and it redirects the random gas outside so it is less likely to hurt people

3 0

3 years ago

What is the approximate terminal velocity of a sky diver before the parachute opens

anzhelika [568]

Answer:

The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.

Explanation:

The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
During terminal velocity, we can represent mathematical equation as;

Buoyancy force + drag force = Gravity

6 0

3 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

4a. The Coyote has an initial velocity vector of $\vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath$ . His position at time $t$ is given by the vector

$\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2$

where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t$

$r_y=15.5\,\mathrm m-\dfrac g2t^2$

The Coyote hits the ground when $r_y=0$ :

$15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s$

4b. Here we evaluate $r_x$ at the time found in (4a).

$r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m$

5. The shell has initial position vector $\vec r_0=(1.52\,\mathrm m)\,\vec\jmath$ , and we're told that after some time the bullet (now separated from the shell) has a position of $\vec r=(3500\,\mathrm m)\,\vec\imath$ .