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Masja [62]
3 years ago
11

What would be the new volume if the pressure of 600mL is increased from 100kPa to 150.56 kPa

Physics
1 answer:
Gelneren [198K]3 years ago
4 0

ASSUMING that the substance is a gas . . .

It all depends on HOW the pressure was increased.

-- IF the gas was left in the same container, and the pressure was increased by heating the container, then the volume hasn't changed. It's still the inside volume of the container.

-- IF the pressure was increased by stuffing all of the gas into a smaller container, then the new volume is the volume of the smaller container.

No matter what the mass or temperature of the gas is, it always expands to fill whatever container you're keeping it in.  

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Technician A says that cabin filters are accessible behind the glove compartment. Technician B says that cabin filters are acces
BartSMP [9]

Answer:

A

but it depends on the car

Explanation:

3 0
3 years ago
A pendulum consists of a 1.7-kg block hanging on a 1.6-m length string. A 0.01-kg bullet moving with a horizontal velocity of 82
Rzqust [24]

Answer:

0.42 m

Explanation:

mass of pendulum, M = 1.7 kg

Length of pendulum , l = 1.6 m

mass of bullet, m = 0.01 kg

initial velocity of bullet, u = 828 m/s

final velocity of bullet, v = 340 m/s

initial velocity of pendulum, U = 0

Let the final velocity of pendulum is V.

Use conservation of momentum for bullet and the pendulum

m x u + M x U = m x v + M x V

0.01 x 828 + 1.7 x 0 = 0.01 x 340 + 1.7 x V

8.28 + 0 = 3.4 = 1.7 V

V = 2.87 m/s

Now the kinetic energy of the pendulum is converted into potential energy of pendulum and let it raised to a height of h from the initial level.

Use energy conservation

Kinetic energy of the pendulum  = potential energy of the pendulum

0.5 x M x V² = M x g x h

0.5 x 2.87 x 2.87 = 9.8 x h

h = 0.42 m

6 0
3 years ago
A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
vodka [1.7K]
We know that tangential acceleration is related with radius and angular acceleration according the following equation:  
at = r * aa  
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)  
So the radius is r = d/2 = 1.2/2 = 0.6 m  
Then at = 0.6 * 5 = 3 m/s2  
Tangential acceleration of a point on the flywheel rim is 3 m/s2
5 0
3 years ago
A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured
nikklg [1K]

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

4 0
3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
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