9.The force of gravity between two asteroids is 10,000 newtons (N).

gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure

Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be ' $\text{P}_2$ '.

As we know that Boyle's law formula states that;

$P_1 \times V_1 = P_2 \times V_2$

where, $P_1$ = original pressure of gas in the container = 310 mm hg

$P_2$ = required new pressure

$V_1$ = volume of gas in the container = 185 ml

$V_2$ = desired new volume of the gas = 74 ml

So, $P_2 = \frac{P_1 \times V_1}{V_2}$

$P_2 = \frac{310 \times 185}{74}$

= 775 mm hg

Hence, the required new pressure is 775 mm hg.

7 0

2 years ago

A current of 0.5 A flows in a 60 W light bulb when the voltage differences between the ends of the filament is 120 V. What is th

LuckyWell [14K]

Resistance = (voltage) / (current)

Resistance = (120 V) / (0.5 A)

Resistance = 240 ohms

Know what ? There might be too much information given in this question. I want to check, because it's possible that it might not even all fit together.

To calculate my answer, I only used the voltage and the current. I didn't use the "60 watts", and I'm curious to know whether it even fits with the given voltage and current.

Power = (voltage) times (current).

Power = (120 V) times (0.5 A)

Power = 60 watts

Well gadzooks and sure enough ! The three numbers given in the question all go together nicely.

And not only THAT !

The answer could have been calculated by using ANY TWO of them.

7 0

2 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

2 years ago

A baseball player throws a ball horizontally at the same moment another baseball player jobs a second ball straight down from th

attashe74 [19]

I don't know what you mean when you say he "jobs" the other ball, and the answer to this question really depends on that word.

I'm going to say that the second player is holding the second ball, and he just opens his fingers and lets the ball drop, at the same time and from the same height as the first ball.

Now I'll go ahead and answer the question that I've just invented:

Strange as it may seem, both balls hit the ground at the same time ... the one that's thrown AND the one that's dropped. The horizontal speed of the thrown ball has no effect on its vertical acceleration, so both balls experience the same vertical behavior.

And here's another example of the exact same thing:

Say you shoot a bullet straight out of a horizontal rifle barrel, AND somebody else drops another bullet at exactly the same time, from a point right next to the end of the rifle barrel. I know this is hard to believe, but both of those bullets hit the ground at the same time too, just like the baseballs ... the bullet that's shot out of the rifle and the one that's dropped from the end of the barrel.

7 0

2 years ago

A 60 Watt light bulb runs for 120 seconds. How much energy does it use?

sattari [20]

ENERGY = POWER X TIME
=60 X 120=7200KWh

6 0

2 years ago