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UNO [17]
4 years ago
11

A weather balloon is inflated to a volume of 27.6 l at a pressure of 736 mmhg and a temperature of 26.1 âc. the balloon rises in

the atmosphere to an altitude where the pressure is 360. mmhg and the temperature is -14.0 âc.
Physics
1 answer:
Arisa [49]4 years ago
8 0
We know from gas equation: 

PV=mRT

Where:
P is pressure
V is the volume of the balloon
m is the mass of gas in the balloon (constant)
R is universal gas constant divided by mean molar wt of air (about 28 g/mol) T is thermodynamic temperature (T in Kelvin; T=273 + t (in deg C)
P1 * V1 =m*R*T1 ---- (i) P2 * V2 =m*R*T2 ---- (ii)
Dividing 1 by 2 we get: 
V1 / V2 = (P2/P1) * (T1/T2)
Given:V1=27.6, P1=736mmhg, T1 = 273+26.1=299.1V2=?, P2 = 360mmhg, T2 = 273+(-14)=259
so, V2 = 27.6*(736/360)*(259/299.1) = 48.86
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A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc
Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

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If u speed up from rest to 12 m/s in 3 seconds, what is your acceleration
irakobra [83]
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4 0
4 years ago
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4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat
nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s

The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s

3 0
3 years ago
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