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ElenaW [278]
3 years ago
6

A spaceship approaches the earth with a speed 0.50c. A passenger in the spaceship measures his heartbeat as 9- beats per minute.

What is his heartbeat rate according to an observer that is at rest relative to the earth? A) 69 beats per minute B) 73 beats per minute C) 65 beats per minute D) 61 beats per minute E) 80 beats per minute
Physics
1 answer:
JulijaS [17]3 years ago
5 0

Answer:

D) 61 beats per minute

Explanation:

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The smallest possible part of an element that can still be identified as the element is called
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Answer:

BLM

Explanation:

ACAB

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Which of these is a source of voltage?
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D, all of these are voltage sources. A battery can power small things, generators can power larger things, and outlets can charge, or power things.
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The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0
avanturin [10]

Answer:

(a) v(t) = 6t^2 - 6t - 12, a(t) = 12t - 6

(b) When 0 \leq t < 0.5, object is slowing down, when t > 0.5 object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

v(t) = \frac{ds(t)}{dt}  = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12

To get the acceleration function, we need to take the derivative of the velocity function.

a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

12t - 6 < 0 \\12t < 6 \\t < 0.5

On the other hand, object is speeding up when a > 0

12t - 6 > 0 \\12t > 6 \\t > 0.5

Therefore, when 0 \leq t < 0.5, object is slowing down, when t > 0.5 object is speeding up.

6 0
3 years ago
A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40
Vinvika [58]

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

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Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

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Time = 3.26 s

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Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

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