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alexandr1967 [171]
2 years ago
9

A baseball is tossed straight up in the air. The table shows the height and velocity of the ball at different times as it moves

through the air.
Time in Air (seconds) Height of Ball Above Ground (feet) Velocity

(feet per second)

0.00 4 32

0.25 11 24

0.50 16 16

0.75 19 8

1.00 20 0

1.25 19 −8

1.50 16 −16

1.75 11 −24

2.00 4 −32

Which statement is true about the kinetic energy of the baseball at the moment it reaches its maximum height?


The baseball's kinetic energy is less than zero at that instant.


At that moment, all the baseball's kinetic energy has been lost as heat to the atmosphere.


The baseball's kinetic energy is greater than zero at that instant.


✅ At that moment, all the baseball's kinetic energy has been converted to potential energy. (correct answer)
Physics
1 answer:
Akimi4 [234]2 years ago
3 0

Answer: At that moment, all the baseball's kinetic energy has been converted to potential energy.

Explanation: I took the test

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An eagle is flying horizontally at a speed of 3.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.10 m
Alinara [238K]

Answer:

10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)

Explanation:

The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

We use the first equation of motion for a free-falling body to obtain v as follows;

v = u + gt....................(1)

where g is acceleration due to gravity taken as 9.8m/s/s

It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

h=ut+gt^2/2.................(2)

Given; h = 6.10m.

since u = 0 for the vertical motion;  equation (2) can be written as follows;

h=\frac{1}{2}gt^2............(3)

substituting;

6.1=\frac{1}{2}*9.8*t^2\\6.1=4.9t^2\\hence\\t^2=6.1/4.9\\t^2=1.24\\t=\sqrt{1.24}=1.12s

Putting this value of t in equation (1) we obtain the following;

v = 0 + 9.8*1.12

v = 10.93m/s

5 0
3 years ago
According to the spectrum seen below, which set of photons were absorbed by the element?
Effectus [21]
The correct choice is ' D ' .

These are the numbers that are missing from the set of blocks at
the top, and also the bars missing from the graph in the middle.

I have no idea what the strip at the bottom labeled "full energy spectrum"
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3 years ago
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In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo
Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144  × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

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An object at free fall near the Earth’s surface experiences constant velocity.
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False it accelerates.
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