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kirza4 [7]
3 years ago
8

3. If a certain quantity of gas has a volume of 30 cm3 at a pressure of 1 × 105 Pa, what is its volume when the pressure is

Physics
1 answer:
arlik [135]3 years ago
5 0

We know that:

                             P_{1}V_{1}= P_{2}V_{2}               .......(1)

Given at P_{1}=30cm^3 , V_{1}=10^5Pa

a)

When P_{2}=2*10^5Pa , V_2=?

Using equation (1)

                                    30*10^5=2*10^5*V_2

                                     V_2=30/2=15cm^3

b)

When P_2=5*10^5Pa, V_2=?

using equation (1)

                                     30*10^5=5*10^5*V_2

                                    V_2=30/5=6cm^3

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Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How
Pachacha [2.7K]

Answer:

The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

Explanation:

Given;

speed of the faster car, v₁ = 60 mi/h

speed of the slower car, v₂ = 55 mi/h

Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles

\frac{x}{55} - \frac{x}{60}   = \frac{15}{60}

Note: divide 15 mins by 60 to convert to hours for consistency in the units.

\frac{x}{55} - \frac{x}{60}   = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles

Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.

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The distance from the Earth to the Sun is 92 868 000 miles.
ANEK [815]
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Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to
lbvjy [14]

Answer:

Explanation:

We shall consider direction towards left as positive Let the required velocity be v and let v makes an angle φ

Applying law of conservation of momentum along direction of original motion

m₁ v₁  - m₂ v₂ = m₂v₃ - m₁ v₄

0.132 x 1.25 - .143 x 1.14 = 1.03 cos43 x .143 - v cos θ

v cos θ = .8

Applying law of conservation of momentum along direction perpendicular to direction of original motion

1.03 sin 43 x .143 = .132 x v sinθ

v sinθ = .76

squaring and adding

v² = .76 ² + .8²

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
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