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3 years ago

A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.

1 answer:

8 0

A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.

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You have a 1.8m long copper wire. You want to make an N-turn current loop that generates a 2.0mT magnetic field at the center wh

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The diameter is 0.022m.

Explanation:

The magnetic field $B$ at the center of the coil is given by

(1). $B = \dfrac{\mu_0 NI}{d}$

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$\therefore N = \dfrac{L}{\pi d }$

putting this into equation (1) we get:

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solve for $d:$

$\boxed{d = \sqrt{\dfrac{\mu_0I L }{\pi B} }}$

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$L= 1.8m$

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4 years ago

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

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Answer:

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$v_A_1-v_B_1=v_{B2}-v_{A2}$

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$v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s$

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