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3 years ago

B) A skilled jet fighter flies a stunt plane in a vertical circle of 1200 ft

Physics

1 answer:

adell [148]3 years ago

6 0

The jet fighter's acceleration at the highest point is 17.56 ft/s²

The jet fighter's acceleration at the lowest point is 35.12 ft/s²

The given parameters;

radius of the circular path, r = 1200 ft

constant velocity, v = 140 mi/h = 205.3 ft/s

The formula for centripetal acceleration in a circular path is given as;

$a_c = \frac{v^2}{r}$

The jet fighter's acceleration at the highest point is calculated as;

h = 2r = 2 x 1200 ft = 2,400 ft

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(205.3)^2}{2400} \\\\a_c = 17.56 \ ft/s^2$

The jet fighter's acceleration at the lowest point is calculated as;

h = r = 1,200 ft

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(205.3)^2}{1200} \\\\a_c = 35.12\ ft/s^2$

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A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

$W=K_f -K_i$

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving,

$K_f = K_i + W=66,120+(-36,733)=29387 J$

The final kinetic energy of the car can be written as

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is its mass

v is its final speed

Solving for v,

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

4 0

4 years ago

OleMash [197]

Answer:

R= 5.4 ohms

Explanation:

Given that

V= 9 V

Power ,P= 15 W

Lets take resistor resistance = R

We know that Power given as

$P=\dfrac{V^2}{R}$

V=Voltage

P=Power

R=Resistance

$R=\dfrac{V^2}{P}$

Now by putting the all the values in the above equation

$R=\dfrac{9^2}{15}$

R= 5.4 ohms

Therefore the resistance of the resister will be 5.4 ohms.

7 0

3 years ago

Steam enters an adiabatic turbine steadily at 7 MPa, 5008C, and 45 m/s, and leaves at 100 kPa and 75 m/s. If the power output of

marusya05 [52]

Answer:

a) $\dot m = 6.878\,\frac{kg}{s}$ , b) $T = 104.3^{\textdegree}C$ , c) $\dot S_{gen} = 11.8\,\frac{kW}{K}$

Explanation:

a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

$-\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The mass flow rate is:

$\dot m = \frac{\dot W_{out}}{h_{in}-h_{out}}$

According to property water tables, specific enthalpies and entropies are:

State 1 - Superheated steam

$P = 7000\,kPa$

$T = 500^{\textdegree}C$

$h = 3411.4\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

State 2s - Liquid-Vapor Mixture

$P = 100\,kPa$

$h = 2467.32\,\frac{kJ}{kg}$

$s = 6.8000\,\frac{kJ}{kg\cdot K}$

$x = 0.908$

The isentropic efficiency is given by the following expression:

$\eta_{s} = \frac{h_{1}-h_{2}}{h_{1}-h_{2s}}$

The real specific enthalpy at outlet is:

$h_{2} = h_{1} - \eta_{s}\cdot (h_{1}-h_{2s})$

$h_{2} = 3411.4\,\frac{kJ}{kg} - 0.77\cdot (3411.4\,\frac{kJ}{kg} - 2467.32\,\frac{kJ}{kg} )$

$h_{2} = 2684.46\,\frac{kJ}{kg}$

State 2 - Superheated Vapor

$P = 100\,kPa$

$T = 104.3^{\textdegree}C$

$h = 2684.46\,\frac{kJ}{kg}$

$s = 7.3829\,\frac{kJ}{kg\cdot K}$

The mass flow rate is:

$\dot m = \frac{5000\,kW}{3411.4\,\frac{kJ}{kg} -2684.46\,\frac{kJ}{kg}}$

$\dot m = 6.878\,\frac{kg}{s}$

b) The temperature at the turbine exit is:

$T = 104.3^{\textdegree}C$

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:

$\dot m \cdot (s_{in}-s_{out}) + \dot S_{gen} = 0$

$\dot S_{gen}=\dot m \cdot (s_{out}-s_{in})$

$\dot S_{gen} = (6.878\,\frac{kg}{s})\cdot (7.3829\,\frac{kJ}{kg\cdot K} - 6.8000\,\frac{kJ}{kg\cdot K} )$

$\dot S_{gen} = 11.8\,\frac{kW}{K}$

4 0

4 years ago

A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the

exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

(√(0.320/K))² = (0.38197)²

0.320/K = 0.1459

K = 2.19 N/m

4 0

4 years ago

Confused as what formulas to used to solve such problem!

Art [367]

Answer:

The trains stop 69 meters apart.

Explanation:

For each train, we're given the initial velocity, final velocity, and acceleration. We want to find the displacement. So we need to use an equation that's independent of time:

v² = v₀² + 2aΔx

First, convert km/h to m/s:

100 km/h = 27.8 m/s

128 km/h = 35.6 m/s

For the first train:

(0 m/s) = (27.8 m/s) + 2 (-0.9 m/s²) Δx

Δx = 429 m

For the second train:

(0 m/s) = (35.6 m/s) + 2 (-0.9 m/s²) Δx

Δx = 702 m

The first train moves 429 m before stopping, and the second train moves 702 m before stopping. They move a total distance of 1131 m. They were initially 1200 m apart, so they stop with 69 m to spare.

7 0

3 years ago