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2 years ago

a jetliner goes from rest to a takeoff speed of 162 mi/h in 39.5 s. what is the magnitude of its average acceleration?

1 answer:

3 0

Acceleration = V/t

First set velocity to m/s

V = 72.42 m/s

Now we divide velocity by time

A = 72.42/39.5
A = 1.83 m/s^2

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An object with a mass of 1500 g (grams) accelerates 10.0 m/s2 when an

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Answer:

15N

Explanation:

F=ma

m=1500g = 1.5kg

a=10m/s2

1.5×10=15 N

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2 years ago

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2 years ago

A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi

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Explanation:

It is given that,

Magnitude of charge, $q=15\ \mu C=15\times 10^{-6}\ C$

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, $B=0.08\ j$

Velocity, $v=(5\ cos25)i+(5\ sin25)j$

$v=[(4.53)i+(2.11)j]\ m/s$

We need to find the magnitude of force on the charge. Magnetic force is given by :

$F=q(v\times B)$

$F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]$

<em>Since</em>, $i\times j=k\ and\ j\times j=0$

$F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]$

$F=0.00000543\ kN$

$F=5.43\times 10^{-6}\ kN$

So, the force acting on the charge is $5.43\times 10^{-6}\ kN$ and is moving in positive z axis. Hence, this is the required solution.

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3 years ago

A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is locat

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Answer:

Q = 12.466μC

Explanation:

For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:

$Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}$

Solving for Q:

$Q = \frac{m*V^{2}*r}{K*q}$

Taking special care of all units, we can calculate the value of the charge:

Q = 12.466μC

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2 years ago

You use 35 J of energy to move an object 5 m. What is the weight of the object

Vlad [161]

Explanation:

35 ÷ 5 = 7kg.

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2 years ago