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3 years ago

7

A man is walking while riding a train. He says he is moving at 2 mph. A woman standing on a platform at a train station says the

man on the train is moving at 72 mph. Which person is correct?

1 answer:

Semenov [28]3 years ago

6 0

The woman on the platform is correct because it is the pace of the man moving on the train not walking.

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An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil

Lilit [14]

Answer:

Explanation:

We shall apply law of conservation of momentum .

m₁ v₁ = m₂ v₂

m₁ = 80 kg

v₁ = ?

m₂ = .43 kg

v₂ = 15 m /s

80 v₁ = .43 x 15

v₁ = .43 x 15 / 80

= .08 m /s

8 cm /s

option A is correct.

Law of conservation of momentum is applied only where no external force acts . Here during the period when ball was thrown , apart from internal force , external force like gravitational force was also acting but we still applied this law because the time duration of action of throw was very small

so we neglected the effect of external force . So it was just an approximation.

7 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

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3 years ago

How do you calculate voltage

IRINA_888 [86]

Voltage=Energy/Charge, V=E/Q V in volts V, E in joules J, Q in coulombs C. or Voltage= Current×Resistance, V=IR V in volts V, I in amperes A and R in ohms Ω

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A wire is placed between the poles of a horseshoe magnet. There is a current in the wire in the direction shown, and this causes

Liula [17]

The answer in the picture

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Which of these is something whose construction MOST LIKELY would be determined by the government? A) An airport B) A golf club C

vesna_86 [32]

The answer is; A

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3 years ago