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notka56 [123]
2 years ago
14

If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w

hat will be the final volume of the sample?
Physics
1 answer:
uranmaximum [27]2 years ago
5 0

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2

V₂ is the final volume

V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{300\times 0.5}{0.75}\\\\V_2=200\ mL

So, the final volume of the sample is 200 mL.

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a roller coaster is at the top of a hill and rolls to the top of a lower hill.If mechanical energy is conserved,on the top of wh
pychu [463]
Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

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7 0
3 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

6 0
3 years ago
Can anybody help me with this Physics question! (DUE TOMORROW) (WILL MARK BRAINLIST)
Ymorist [56]
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A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up
prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,  

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)  

We know that, Net force = m × a

\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}

\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}

\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}

a_{\text {rad }}=2.45 \mathrm{m} / \mathrm{s}^{2}

Now, Net force = m × a

Net force = 790 × 2.45

Net force = 1935.5 N

4 0
3 years ago
What is the acceleration of a 10kg pushed by a 5n force
Semenov [28]
Force equals mass times acceleration. Or:
F=ma
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8 0
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