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ruslelena [56]
4 years ago
14

A 12 oz can of soda was left in a car on a hot day. in the morning, the soda temperature was 60°f with a gauge pressure of 40 ps

i. what will the gauge pressure be when the soda reaches 90°f (or 305 kelvin)?
Physics
1 answer:
enot [183]4 years ago
4 0
From gas equations
PV/T = Constant

Then
P1V1/T1 = P2V2/T2, but V1 = V2

Therefore,
P1/T1 = P2/T2
P2 = (P1T2)/T1

P1 = 40 psi
T1 = 60°F = 288.706 K
T2 = 90°F = 305.372 K

Substituting;
P2 = (40*305.372)/288.706 = 42.31 psi

Then, gauge pressure would read 42.31 psi.
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You can use the displacement method or the eureka can so basically in the displacement can what you have to do is to put some water into a measuring cylinder and measure its volume before adding the irregular shaped object and then measuring the level of water which had been displaced and then eureka can you can check online
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Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?
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Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity. 
4 0
3 years ago
Read 2 more answers
Puck 1 (1 kg) travels with velocity 30 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After th
Katyanochek1 [597]

Answer:

v₂ = 22.5 m/s

Explanation:

Given that

For puck 1

m₁= 1 kg

u₁= 30 m/s

For puck 2

m₂= 1 kg

u₂= 0 m/s

After collision

Puck 1 have velocity v₁=7.5 m/s

Take puck 2 will have velocity v₂

From linear momentum conservation

P₁=P₂

m₁ u₁+m₂ u₂=m₁ v₁+m₂ v₂

1 x 30 + 1 x 0 = 1 x 7.5 + 1 x v₂

30 - 7.5 =v₂

v₂ = 22.5 m/s

6 0
3 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
A child blows a leaf from rest straight up in the air. The leaf has a constant upward acceleration of magnitude 1.0\,\dfrac{\tex
Novosadov [1.4K]

Answer:

Explanation:

Given

Acceleration a = 1.0m/s²

Displacement S = 1.0m

Required

Time t taken by the leaf to displace

Using the equation of motion

S = ut+1/2at²

Substitute

1.0 = 0+1/2(1)t²

1 = t²/2

Cross multiply

t² = 2

t = ±√2

t = 1.41secs

It takes the leaf to 1.41s to displace by 1m upward

6 0
3 years ago
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