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Hunter-Best [27]
3 years ago
14

John runs around a 126.5 m circular track 3.5 times in 4.17 minutes. What is his average speed?

Physics
1 answer:
sdas [7]3 years ago
6 0
Average speed = distance traveled / time

average speed  = (126.5 m * 3.5 laps) / (4.17 min)

= 106.2 m/min
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3 years ago
A photon with wavelength λ = 0.0830 nm is incident on an electron that is initially at rest. if the photon scatters in the backw
QveST [7]
Compton scattering equation of wavelengths 
λ'-λ = h/mec (1 - CosФ)
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7 0
3 years ago
What is the speed of an object that moves from 70 m to 68 m in 14 s?
PtichkaEL [24]
Speed = m/s
Velocity can be negative, but speed cannot.

70 - 68 = 2
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5 0
2 years ago
Write examples of adaptation in mangrove .? <br>what are the adaptation of rainforest.?​
Sergio039 [100]

Answer:

example of adaptation in mangrove

Mangroves are salt-tolerant trees, also called halophytes, and are adapted to life in harsh coastal conditions. They contain a complex salt filtration system and complex root system to cope with salt water immersion and wave action. They are adapted to the low oxygen conditions of waterlogged mud.

adaptation of rainforest

Many animals have adapted to the unique conditions of the tropical rainforests.

hope this can help you

6 0
3 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

5 0
3 years ago
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