What is a physical change ?

A tin can has a radius of 0.0410m and is 185 m high if the air is removed from inside the can how much forces will the outside a

grin007 [14]

The force exerted on the bottle is 532 N

Explanation:

When the air is removed from inside the bottle, vacuum is created, therefore the external pressure (the atmospheric pressure) is no longer balanced, and it creates a net force downward on the can.

The pressure is related to the force by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area of the can

Here we have:

$p=1.01\cdot 10^5 Pa$ is the atmospheric pressure

r = 0.0410 m is the radius of the can, so the area is

$A=\pi r^2 = \pi (0.0410)^2=5.27\cdot 10^{-3}m^2$

And solving for F, we can find the force on the can:

$F=pA=(1.01\cdot 10^5)(5.27\cdot 10^{-3})=532 N$

Learn more about pressure:

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3 0

4 years ago

If one motor has three times as much power as another, then the smaller power motor: A. can do the same work in the same time. B

Dvinal [7]

Answer:B

Explanation:

Given

First motor has three times as much as power as another

Let the power of smaller motor be P

Therefore bigger motor has power 3 P

and we know that energy $E=power\times time$

Larger motor $E=3P\times t=3Pt$

Smaller motor $E=P\times 3t=3Pt$

Therefore smaller motor will do same work if it takes three times the time taken by Larger motor

8 0

4 years ago

Two boxes (m1 = 43.0 kg and m2 = 39.5 kg) are connected by a light string that passes over a light, frictionless pulley. One box

wlad13 [49]

The net force on the system:
F = m₂g - m₁gsin(∅)
F = 39.5 x 9.81 - 43 x 9.81 x sin(30)
F = 176.58 N

Now, we use F = ma to find the acceleration on each mass.
F = m₁a₁
a₁ = 176.58 / 43
a₁ = 4.11 m/s²

F = m₂a₂
a₂ = 176.58 / 39.5
a₂ = 4.47 m/s²

5 0

4 years ago

A spotlight on the ground is shining on a wall 24m away. If a woman 2m tall walks from the spotlight toward the building at a sp

Lubov Fominskaja [6]

Answer:

$\dfrac{dy}{dt}=-0.059\ m/s$

Explanation:

It is given that,

Distance between the spotlight and the wall, y = 24 m

Height of the woman, h = 2 m

The woman walks toward the building at the rate of 0.6 m/s, $\dfrac{dx}{dt}=0.6\ m/s$

In the attached figure, triangle ABC and MNC are similar. So,

$\dfrac{2}{y}=\dfrac{x}{24}$ ............(1)

$y=\dfrac{48}{x}$

When she is 2 meters from the building. So x = 24-2 = 22 m

$y=\dfrac{48}{22}=2.18\ m$

Differentiating equation (1) i.e.

$xy=48$

$x.\dfrac{dy}{dt}+y.\dfrac{dx}{dt}=0$

$22.\dfrac{dy}{dt}+2.18\times 0.6=0$

$\dfrac{dy}{dt}=-0.059\ m/s$

So, her shadow is decreasing at the rate of 0.059 m/s. Hence, this is the required solution.

7 0

4 years ago

A car leaves a garage at 10 mph. it speeds up to 20 miles per hour in the city and eventually increases to 50 miles per hour in

LekaFEV [45]

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

= (50 miles) / (1.5 hours)

= (50/1.5) miles/hours

= (33 and 1/3) mph .

We don't care about all that other "data" given earlier in the question.
We only need to know the total distance covered and the time it took
to cover the distance.

4 0

3 years ago