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bixtya [17]
1 year ago
13

A ball is thrown vertically upward with an initial speed of 50 m/s. if air resistance does affect motion, then what would be its

speed upon returning to its starting point?
Physics
1 answer:
olganol [36]1 year ago
8 0

Answer:

your answer would 50 m/s I do believe

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Which of the following is not an example of natural climate change?
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(c) is the correct choice.

El Nino (a), Earth's orbit (b), and solar energy output (d) are all "natural" occurrences.  You can't do a thing aboutum.

Fossil fuels ... or, more precisely, humanity's use of vast quantities of fossil fuels as a convenient source of huge quantities of energy ... and the subsequent increase of Carbon Dioxide in the planet's atmosphere, is not the result of "natural" processes.  It's the result of human efforts to <em>alter and control</em> Nature, through <em>artificial</em> processes.

7 0
2 years ago
In a parallel circuit, if one connection is broken
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C hope that helps hens
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A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find
garik1379 [7]
Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph

The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.

Answer:
The magnitude is 18.1 mph.
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8 0
2 years ago
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5 0
2 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0
2 years ago
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