Find the mass of a 165 N child

How is humidity related to air pressure?

Vlad [161]

Humid air has higher pressure because of the heaviness of the water

7 0

3 years ago

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A fairground ride consists of a large vertical drum that spins so

expeople1 [14]

Answer:

v = 3.84 m/s

Explanation:

In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:

$Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}$

where,

v = minimum speed = ?

g = acceleration due to gravity = 9.81 m/s²

r = radius = 10 m

μ = coefficient of friction = 0.15

Therefore,

$v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}$

6 0

3 years ago

Humans evolved in Earth's atmosphere, therefore the pressures interior to the body are relatively close to atmospheric pressure.

harina [27]

Answer:

194516 sheets

Explanation:

So the area of each sheet of paper is:

A = 0.216 * 0.279 = 0.060264 square meters

For the paper sheet to make the same effect as the atmospheric pressure P, then the gravity F from the paper sheet must be

F = AP = 0.060264 * 101325 = 6106 N

Let g = 9.81 m/s2, then the mass of paper needed to generate that gravity is

m = F/g = 6106 / 9.81 = 622.4 kg

If each sheet has a mass of 0.0032 kg, then the total number of sheets to have that much mass is

622.4 / 0.0032 = 194516 sheets

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3 years ago

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The imaginary line from the tip of the football to the football to the sideline is called

pentagon [3]

The imaginary line from the tip of the football to the football to the sideline is called Line of Scrimmage. I believe.

8 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

$\Delta U = q \Delta V$

where

q is the charge transferred

$\Delta V$ is the potential difference between the ground and the clouds

Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

$K=\frac{1}{2}mv^2$

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

5 0

3 years ago