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2 years ago

8

Help please? I don’t know this

1 answer:

ohaa [14]2 years ago

4 0

Answer:

Gas giants rotate faster than those of earth .

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A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all

timurjin [86]

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, $a=5\times 10^{-4}\ m^2$

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}$

$P=1.4\times 10^5\ Pa$

We know that the atmospheric pressure is given by :

$P_o=1.013\times 10^5\ Pa$

So, the pressure is given by :

$p=P+P_o$

$p=1.4\times 10^5+1.013\times 10^5$

$p=2.41\times 10^5\ Pa$

Hence, this is the required solution.

7 0

2 years ago

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/

larisa [96]

Answer:

Total momentum, p = 21.24 kg-m/s

Explanation:

Given that,

Mass of first piece, $m_1=200\ g= 0.2\ kg$

Mass of the second piece, $m_2=300\ g= 0.3\ kg$

Speed of the first piece, $v_1=82\ m/s$ (along x axis)

Speed of the second piece, $v_2=45\ m/s$ (along y axis)

To find,

The total momentum of the two pieces.

Solve,

The total momentum of two pieces is equal to the sum of momentum along x axis and along y axis.

$p_x=m_1v_1$

$p_x=0.2\ kg\times 82\ m/s$

$p_x=16.4\ kg-m/s$

$p_y=m_2v_2$

$p_y=0.3\ kg\times 45\ m/s$

$p_y=13.5\ kg-m/s$

The net momentum is given by :

$p=\sqrt{p_x^2+p_y^2}$

$p=\sqrt{16.4^2+13.5^2}$

p = 21.24 kg-m/s

Therefore, the total momentum of the two pieces is 21.24 kg-m/s.

4 0

2 years ago

At t=0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a con

fiasKO [112]

Answer:

e.26m/s

Explanation:

Vf=Vi+at (1)

Vf=9j+(2i-4j)t

X= X₀+at

now, in the i direction

15=O+2t or t=7.5 when x position is 15

Lets put that into the (1) equation, solve for Vf.

Vf=9j+(2i-4j)7.5

Vf= 15i - 21j

Speed= $\sqrt{xcomponent^2 + ycomponent^2}$

Vf= 25.8 m/s

5 0

2 years ago

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

7 0

3 years ago

A patient visited a doctor and complained that from the last 2 months, his urine output is very low. After examination, the doct

Dvinal [7]

Answer:

kidney

Explanation:

thats where your pee goes through

8 0

1 year ago

Read 2 more answers