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Margaret [11]
3 years ago
7

A scientist plans to find out the percent of teenagers who like science. She interviews 500 teenagers leaving a science museum a

nd finds that 450 of them like science. The scientist concludes that 90 percent of teenagers like science. Why is the scientist's conclusion most likely unreliable? adults were not interviewed the investigation was time consuming the source of information was biased very few teenagers were interviewed
Physics
2 answers:
creativ13 [48]3 years ago
7 0
Sadly she has not concluded much. She needs a more diverse & larger population of teenagers to come to a less biased conclusion. results will be skewed.

Georgia [21]3 years ago
5 0
The source of information was biased. It was like walking along a river bank in the country and asking everybody you meet whether they like fishing. Or asking 500 people sitting in the bleachers whether they like baseball.
I'm sure the scientist would have gotten different data if she interviewed 500 teenagers at neighborhood basketball courts, or 500 teenagers at a rock concert.
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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

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Answer: The motions shifts.

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As we know,

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E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

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E=ρ*b* (b-a)*/r*εo

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