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Ne4ueva [31]
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?
Physics
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
AveGali [126]

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

U_{gravity} = U_{spring}\\ mgh = \frac{1}{2} kh^{2}

Solving for k:

k = 2mg\frac{1}{h}

5 0
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.

3 0
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.

Learn more here:brainly.com/question/24666225

4 0
2 years ago
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