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3 years ago

A 1200-kg car moving at 20 km/h is accelerated

Engineering

1 answer:

Grace [21]3 years ago

4 0

Answer:

Force on the car will be 4800 N and time required to cover this distance 13.75 sec

Explanation:

We have given mass of the car = 1200 kg

Initial velocity u = 20 km/h

Final velocity v = 75 km/h

Acceleration $a=4m/sec^2$

From the first equation of motion we know that

v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So $75=20+4\times t$

t = 13.75 sec

From second law of motion we know that $F=ma$

So force $F=1200\times 4=4800N$

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What could happen in the aviation

Alina [70]

Answer:

Yes

Explanation:

7 0

3 years ago

A gear motor can develop 2 hp when it turns at 450rpm. If the motor turns a solid shaft with a diameter of 1 in., determine the

kramer

Answer:

Maximum shear stress is;

τ_max = 1427.12 psi

Explanation:

We are given;

Power = 2 HP = 2 × 746 Watts = 1492 W

Angular speed;ω = 450 rev/min = 450 × 2π/60 rad/s = 47.124 rad/s

Diameter;d = 1 in

We know that; power = shear stress × angular speed

So,

P = τω

τ = P/ω

τ = 1492/47.124

τ = 31.66 N.m

Converting this to lb.in, we have;

τ = 280.2146 lb.in

Maximum shear stress is given by the formula;

τ_max = (τ•d/2)/J

J is polar moment of inertia given by the formula; J = πd⁴/32

So,

τ_max = (τ•d/2)/(πd⁴/32)

This reduces to;

τ_max = (16τ)/(πd³)

Plugging in values;

τ_max = (16 × 280.2146)/((π×1³)

τ_max = 1427.12 psi

7 0

3 years ago

What is the value of the work interaction in this process?

Cloud [144]

Answer:

The answer is " $-121\ \frac{KJ}{Kg}$ ".

Explanation:

Please find the correct question in the attachment file.

using formula:

$\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \ or \ \ \frac{RT_1 -RT_2}{n-1}\\\\$

$W =\frac{R(T_1 -T_2)}{n-1}\\\\$

$=\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\$

7 0

3 years ago

Una frase de: ama la vida quien___________________________________

TEA [102]

Answer:

A phrase from: who loves life

Explanation:

5 0

3 years ago

Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat

Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 = 0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = $\frac{m}{\rho}$

V = Av = $\frac{\pi}{4} * d^2 * v1$

V = $\frac{\pi}{4} * 0.28^2 * 5$

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = $\frac{V}{v}$ = $\frac{0.30787}{0.11418}$

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = $\frac{v2}{A2}$

v2 = $\frac{0.3705}{\frac{\pi}{4} * 0.28^2}$

v2 = 6.017 m/s

7 0

3 years ago