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oee [108]
3 years ago
8

A cyclist has a constant speed of 12 m/s. What is the magnitude of the displacement of the cyclist after 18 seconds?

Physics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

216 m

Explanation:

Assuming a straight line:

Δx = vt

Δx = (12 m/s) (18 s)

Δx = 216 m

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The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic
Shkiper50 [21]

Answer:

i) 0.7

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iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight =   \frac{2uSin(A)}{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = \frac{2(4)Sin(60)}{10} = 0.7

ii) distance travel = Range =  R = \frac{u^{2}Sin(2A) }{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = \frac{4^{2}Sin(2*60) }{10} = 1.39

iii) Maximum Height = H = \frac{u^{2}(Sin(A))^{2}  }{2g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: \frac{4^{2}(Sin60)^{2}  }{2*10} = 0.6

4 0
3 years ago
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer
son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$  rev/sec

                             $=\frac{2 \pi \times 7.5}{8}$  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

   $$=1640 \times (5.89)^2 \times 7.5  

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Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

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If toner particles in a laser printer have a negative charge, then what charge do you think the surface of the paper in the prin
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Answer:

Explanation:

As it moves along, the paper is given a strong negative electrical charge by another corona wire. When the paper moves near the drum, its negative charge attracts the positively charged toner particles away from the drum.

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Answer:

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{<em>refer to the above attachment}</em>

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