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Georgia [21]
2 years ago
11

A 2200-kg railway freight car coasts at 4.1 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

r. If the speed of the loaded freight car must not go below 3.4 m/s, what is the maximum mass of grain that it can accept?
Physics
1 answer:
777dan777 [17]2 years ago
8 0

Answer:

The answer is "2.41 \times 10^3"

Explanation:

Given:  

m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\

Using formula:  

\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}

p_i, p_f = system initial and final linear momentum.

V_i, v_f = system original and final linear pace.

m_i = original weight of the car freight.

m_f= car's maximum weight

= \frac{ 2000 \times 4.1}{3.4}\\\\= \frac{ 8.2\times 10^3}{3.4}\\\\= 2.41 \times 10^3

\boxed{m_f = 2.41 \times 10^3}

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An airplane has wings, each with area A, designed so that air flows over the top of the wing at 265 m/s and underneath the wing
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Answer

given,

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speed of underneath wings = 234 m/s

mass of the airplane =  7.2 × 10³ kg

density of air =  1.29 kg/m³

using Bernoulli's equation

 P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2

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3 years ago
A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

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First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

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Now put all the given value in the above formula, we get:

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Now we have to calculate the change in internal energy of the system.

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As we know that, work done is zero at constant volume. So,

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