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Elena-2011 [213]
2 years ago
9

A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that fina

l temperature is 97 "C. The specific heats are Cp 1.005 kJ(kg.K) and C-0.718 kJ/(kg.K). Determine the total work done in kJ:
Physics
1 answer:
lisov135 [29]2 years ago
5 0

Answer:

-58.876 kJ

Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

⇒W = -mCvΔT

⇒Q = -1×0.718×(97-15)

⇒Q = -58.716 kJ

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The three resistors are connected to the same points of the circuit, so they are in parallel configuration. The equivalent resistance of 3 resistors in parallel is given by:
\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}
If we plug the values of the resistances into the formula, we find
\frac{1}{R_{eq}}= \frac{1}{3.0 \Omega}+ \frac{1}{6.0 \Omega}+ \frac{1}{9.0 \Omega}= \frac{6+3+2}{18.0 \Omega} = \frac{11}{18.0 \Omega}
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3 years ago
A bowling ball of mass 5.8 kg moves in a
evablogger [386]

Answer:

Velocity of the ping pong ball must be = V2= 6,035.34m/s

Explanation:

M1= momentum of the bowling ball

m1 = mass of the bowling ball= 5.8kg

v1= velocity of the bowling ball= 1.59m/s

M2= momentum of the ping pong ball

m2= mass of the ping pong ball= 1.528 g/1000=  0.001528kg

v2= velocity of the ping pong ball

Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s

Momentum of the ping pong ball = M2= M1= m2v2

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3 years ago
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet
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Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

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Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

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the answer is static friction


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How do you calculate acceleration
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A=f/m
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A=5
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