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andreyandreev [35.5K]
3 years ago
8

Suppose there is a sample of xenon in a rectangular container. The gas exerts a total force of 4.47 N perpendicular to one of th

e container walls, whose dimensions are 0.125 m by 0.209 m . Calculate the pressure of the sample.
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

The pressure is P=170.61 Nm^{-2}.

Explanation:

Calculate the area of the rectangular container.

A=LB

Here, L is the length and B is the breadth.

Put L=0.125 m  and B=  0.209 m.

A=(0.125)(0.209)

A = 0.0262 m^{2}

The expression for the pressure in terms of area and force is as follows;

P=\frac{F}{A}

Here, P is the pressure and F is the force.

Put  A = 0.0262 m^{2} and F= 4.47 N.

P=\frac{4.47}{0.0262}

P=170.61 Nm^{-2}

Therefore, the pressure of the given sample is P=170.61 Nm^{-2}.

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The radius of Earth is about 6450 km. A 7070 N spacecraft travels away from Earth. What is the weight of the spacecraft at a hei
Triss [41]

Answer:

(a) 1767.43 N

(b) 182.45 N

Explanation:

Radius of earth, R = 6450 km

Weight of person, W = 7070 N

mass of person, m = W / g = 7070 / 9.8 = 721.4 kg

(a) h = 6450 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+6450} \right )^2

g' = g / 4 = 9.8 / 4 = 2.45 m/s^2

The weight of the person at such height is

W' = m x g' = 721.4 x 2.45

W' = 1767.43 N

(b) h = 33700 km

The value of acceleration due to gravity on height is given by

g' = g\left ( \frac{R}{R+h} \right )^2

g' = g\left ( \frac{6450}{6450+33700} \right )^2

g' = g x 0.0258 = 9.8 x 0.0258 = 0.253 m/s^2

The weight of the person at such height is

W' = m x g'

W' = 721.4 x 0.253

W' = 182.45 N

3 0
3 years ago
During its lifespan, what characteristics of the sun will change
4vir4ik [10]
During its lifepsan, the sun's core would keep contracting and heating up.

The temperature will keep increasing to the point where the temperature outside the core will get to hydrogen fusion temperatures.
The sun will grow in surface and eventually became the Red Giant
6 0
3 years ago
Read 2 more answers
Grindstone: Rotational Dynamics and Kinematics You have a grindstone (a disk) that is 133.0 kg, has a 0.635 m radius, and is tur
Colt1911 [192]

Answer:

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

where <em>N</em> can be obtained as follows:

∑ Fc = m*ac   ⇒   N - F = m*ac = m*ω²*R    ⇒  N = F + m*ω²*R

then if

F = 32 N

m = 133 Kg

R = 0.635 m

ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s

we get

N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N

Finally

Ff = μ*N = 0.10*(8390.53 N) = 839.05 N

3 0
3 years ago
If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula
snow_lady [41]

Here we know that

\omega_i = - 8.4 rad/s

\alpha = - 2.8 rad/s^2

t = 1.5 s

now from kinematics we have

\omega_f = \omega_i + \alpha t

now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

\omega_f = -8.4 + (-4.2)

\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

6 0
3 years ago
Select the correct answer.
spin [16.1K]

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

7 0
3 years ago
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