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Bad White [126]
3 years ago
10

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

the required takeoff velocity at the end of the runway. What is the time tTO needed to take off? Express your answer in seconds using three significant figures.
Physics
1 answer:
tiny-mole [99]3 years ago
3 0

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

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erma4kov [3.2K]

Explanation:

Given that,

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Number of turns in the coil, N = 500

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B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^4

Resistance of the coil, R = 640 ohms

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\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=N\pi r^2\dfrac{-dB}{dt}\\\\\epsilon=N\pi r^2\times \dfrac{-d(1.2\times 10^{-2}t+2.6\times 10^{-5}t^4)}{dt}\\\\\epsilon=N\pi r^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=500\pi \times (4.2\times 10^{-2})^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=2.77(1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\ V

Hence, this is the required solution.

4 0
3 years ago
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A tow rope pulls a 1450 kg truck, giving it an acceleration 1.25 m/s^2. What force does the rope exert?
bagirrra123 [75]

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4 0
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Oil explorers set off explosives to make loud sounds, then listen for the echoes from underground oil deposits. Geologists suspe
Elenna [48]

Answer:

dg= 942m

Explanation:

given the depth of the granite Us dg = 500m

time between the explosion t = 0.99s

the speed of sound in granite is Vg = 6000m/s

First of all calculate the time it takes the sound waves to travel down through the lake

Vw = dw/t1

t1 = dw/Vw

t1 = 500/1480

t1 = 0.338s.

Let dg be the depth of the granite basin, so the time it takes for the sound to travel down through the granite is t2 = dg/6000m/s......equation(1)

So the total time it takes to travel down to the oil surface will be

t1/2 = t1 + t2

t1/2=  0.338 + dg/6000.

since the reflection on the oil does not change the speed of sound, the sound will take travelling upto the surface the same time it takes to reach the oil

so; t = 2 t1/2

t1/2 = t/2 = 0.99s/2 = 0.495

Now insert into the values of t1/2 into the equation (1) and solve for dg;

we get 0.495 = 0.338 + dg/6000

dg = (0.495 - 0.338) x 6000

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3 0
3 years ago
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Energy is measured in watts and work is measured in joules. true or false?
barxatty [35]
It's true
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3 0
4 years ago
a body of mass 20kg initially at rest is subjected to a force of 40N for 1sec calculate the change in kinetic energy showing the
Tema [17]

Answer:

Change in KE is 40 J

Explanation:

Recall that the impulse exerted on an object equal the change of momentum of the object (ΔP), which in time is defined as the product of the force exerted on it times the time the force was acting:

Change in momentum is:   ΔP = F * Δt

In our case,

ΔP = 40 N * 1 sec = 40 N s

Since the object was initially at rest, its initial momentum was zero, and the final momentum should then be 40 N s.

So, the initial KE was 0, and the final (KEf) can be calculated using:

KEf = 1 /(2 m) Pf^2 = 1 / (40) 40^2 = 40 J

So, the change in kinetic energy is:

KEf - KEi = 40 J - 0 j = 40 J

8 0
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