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3 years ago

5

An airplane, starting from rest, moves down the runway at constant acceleration for 18 s and then takes off at a speed of 60 m/s

. What is the average acceleration of the plane?

2 answers:

ruslelena [56]3 years ago

4 0

Answer:

a =3.33 m/s²

Explanation:

given,

initial speed of Plane, u = 0 m/s

final speed of plane, v = 60 m/s

time of the acceleration, t = 18 s

average acceleration of the plane, a = ?

average acceleration is equal to change in velocity per unit time.

$a = \dfrac{v - u}{t}$

$a = \dfrac{60 - 0}{18}$

$a = \dfrac{60}{18}$

a =3.33 m/s²

Hence, average acceleration of the plane is equal to a =3.33 m/s²

ElenaW [278]3 years ago

3 0

Answer:

Explanation:

Given

Airplane runs for a time period of $t=18\ s$

Airplane takes off with a velocity of $v=60\ m/s$

It is given that airplane starts from rest i.e. initial velocity $u=0$

acceleration is the rate of change of velocity thus it is given by

$a=\frac{v-u}{t}$

$a=\frac{60-0}{18}$

$a=3.33\ m/s^2$

Average acceleration for 18 s is $3.33 m/s^2$

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

Tcecarenko [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

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Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

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Julli [10]

Answer:

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Explanation:

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