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3 years ago

14

An 85 kg object is moving at a constant speed of 15 m/s in a circular path, which has a radius of 20 meters. What centripetal fo

rce is exerted on the body?

1 answer:

ale4655 [162]3 years ago

5 0

The centripetal force is equal to:
$\frac{mv^2}{r}$
After substituting the values from the question into the equation, we get:
<span> $\frac{85*15^2}{20}$
</span>This solves to give a value of 956.25N

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A 50-cm-long spring is suspended from the ceiling. A 270 g mass is connected to the end and held at rest with the spring unstret

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Answer:

k = 22.05 N/m

Explanation:

The potential energy of the mass is converted into potential energy of the spring.

Given:

mass m = 0.27 kg

gravitational constant g = 9.8 m/s²

distance falling/ stretching of spring h = 0.24 m

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3 years ago

⦁ A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the c

Rudik [331]

Answer:

The value of temperature at compressor outlet = 543.43 K

The temperature at turbine outlet = 773.04 K

Back work ratio = 0.388

The value of efficiency of the Ideal Brayton cycle =0.448

Explanation:

Pressure ratio ( $r_{p}$ )= 8 And specific heat ratio ( $\gamma$ ) = 1.4

Compressor inlet temperature ( $T_{1}$ ) = 300 K

Turbine inlet temperature ( $T_{3}$ ) = 1400 K

(a). Temperature ratio inside the compressor is given by

$\frac{T_{2} }{T_{1} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{1}$ & $T_{2}$ represents the compressor inlet and outlet temperature.

Put all the values in the given formula we get

⇒ $\frac{T_{2} }{300}$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $\frac{T_{2} }{300}$ = 1.811

⇒ $T_{2}$ = 543.43 K

This is the value of temperature at compressor outlet.

The temperature ratio inside the turbine is given by

$\frac{T_{3} }{T_{4} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{3}$ & $T_{4}$ represents the turbine inlet & outlet temperatures.

Put all the values in the given formula we get

⇒ $\frac{1400}{T_{4} }$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $T_{4}$ = 773.05 K

This is the temperature at turbine outlet.

(b). The work done by the turbine is given by the formula ( $W_{T}$ ) = $C_{p}$ ( $T_{3}$ - $T_{4}$ )

Put all the values in the above formula we get $W_{T}$ = 1.005 × (1400 - 773.05)

$W_{T}$ = 630.08 $\frac{KJ}{Kg}$

And work done inside the pump is given by $W_{c}$ = $C_{p}$ ( $T_{2}$ - $T_{1}$ )

Put all the values in the above formula we get $W_{c}$ = 1.005 × (543.43 - 300)

$W_{c}$ = 244.65 $\frac{KJ}{Kg}$

Back work ratio is given by $r_{b}$ = $\frac{W_{C} }{W_{}T}$

Put the values of $W_{c}$ & $W_{T}$ in above formula we get

⇒ $r_{b}$ = $\frac{244.65}{630.08}$

⇒ $r_{b}$ = 0.388

This is the value of back work ratio.

(C). Thermal Efficiency is given by E = 1 - $\frac{1}{r_{p} ^{\frac{\gamma - 1}{\gamma} } }$

⇒ E = 1 - $\frac{1}{8 ^{\frac{1.4 - 1}{1.4} } }$

⇒ E = 0.448

This is the value of efficiency of the Ideal Brayton cycle.

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3 years ago

28. A car with an initial velocity of 24.5 m/s east has an

zlopas [31]

The displacement of the car after the given final velocity is 31.59 m.

The given parameters;

Initial velocity of the car, u = 24.5 m/s east
Final velocity of the car, v = 18.3 m/s east
Acceleration of the car, a = 4.2 m/s²

The displacement of the car after the given final velocity is obtained by applying the third kinematic equation as shown below;

v² = u² + 2as

$s = \frac{v^2 - u^2}{2a} \\\\s = \frac{(18.3)^2 - (24.5)^2}{2(-4.2)} \\\\s = 31.59 \ m$

Thus, the displacement of the car after the given final velocity is 31.59 m.

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2 years ago