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Makovka662 [10]
4 years ago
14

A jet plane is cruising at 310 m/s when suddenly the pilot turns the engines up to full throttle. after traveling 4.0 km , the j

et is moving with a speed of 400 m/s.
Physics
1 answer:
Inessa [10]4 years ago
4 0
For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad 
where:
Vf = 400m/s 
Vi = 300m/s 
a = ? 
d = 4.0km 
= 4000m 

400^2 = 300^2 + 2a4000 
a = [ 160000 - 90000 ] / 8000 
a = 8.75m/s^2 
rounding it off to 2 significant figures, will give us 8.8 m/s^2.
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A freight train consists of two 8.00 \times 10^5~\text{kg}8.00×10 ​5 ​​ kg engines and 45 cars with average masses of 5.50 \time
erma4kov [3.2K]

Answer:

F₁ = 4.29 x 10⁵ N

Explanation:

The total force required to move the freight train with the given acceleration is given by the following formula:

F = ma + f

where,

F = Total Force Required from both engines = ?

m = equivalent mass of system = 2(8 x 10⁵ kg) + 5.5 x 10⁵ kg = 21.5 x 10⁵ kg

a = required acceleration = 5 x 10⁻² m/s²

f = force of friction = 7.5 x 10⁵ N

Therefore,

F = (21.5 x 10⁵ kg)(0.05 m/s²) + 7.5 x 10⁵ N

F = 8.575 x 10⁵ N

Now, for identical forces in each engine can be given as:

Force exerted by each engine = F₁ = F/2

F₁ = 8.575 x 10⁵ N/2

<u>F₁ = 4.29 x 10⁵ N</u>

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A solenoid having an inductance of 6.95 μh is connected in series with a 1.24 kω resistor. (a) if a 12.0 v battery is connected
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In electrical circuit, this arrangement is called a R-L series circuit. It is a circuit containing elements of an inductor (L) and a resistor (R). Inductance is expressed in units of Henry while resistance is expressed in units of ohms. The relationship between these values is called the impedance, denoted as Z. Its equation is

Z = √(R^2 + L^2)
Z =  √((1.24×10^3 ohms)^2 + (6.95×10^-6 H)^2)
Z = 1,240 ohms

The unit for impedance is also ohms. Since the circuit is in series, the voltage across the inductor and the resistor are additive which is equal to 12 V. Knowing the impedance and the voltage, we can determine the maximum current.
I = V/Z=12/1,240 = 9.68 mA
But since we only want to reach 73.6% of its value, I = 9.68*0.736 = 7.12 mA. Then, the equation for R-L circuits is
I= \frac{V( 1- e^{-t/τ}  )}{R}, where τ = L/R = 6.95×10^-6/1.24×10^3 = 5.6 x 10^-9
Then,
7.12x 10^{-3} = \frac{12( 1- e^{-t/5.6x 10^{-9} } )}{1240}

t = 7.45 nanoseconds
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