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Vladimir [108]
3 years ago
5

Which of the following is NOT an effective pre-exercise energizing snack?

Physics
2 answers:
maks197457 [2]3 years ago
4 0
Cookies would not be an effective pre-exercise energizing snack, as their high sugar content would be absorbed quickly by the body, not sustaining the athlete for the duration of their event.
Rom4ik [11]3 years ago
4 0
The answer is cookies
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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release.
Alex_Xolod [135]

(a) The  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

(b) The velocity of the ball when it reaches its maximum height is zero.

(c) The initial velocity of the ball is 17.36 m/s.

(d) The maximum height it reaches is 15.36 m.

<h3>Acceleration of the ball</h3>

The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

<h3>Velocity of the ball at maximum height</h3>

The velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.

<h3>Initial velocity of the ball</h3>

v = u - gt

at maximum height, final velocity, v = 0

0 = u - gt

u = gt

u = 9.81 x 1.77

u = 17.36 m/s

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

h = ut - ¹/₂gt

h = 17.36(1.77) - ¹/₂(9.81)(1.77²)

h = 15.36 m

Thus, the  acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.

The velocity of the ball when it reaches its maximum height is zero.

The initial velocity of the ball is 17.36 m/s.

The maximum height it reaches is 15.36 m.

Learn more about maximum height here: brainly.com/question/12446886

#SPJ1

3 0
2 years ago
What is the terminal velocity of blood? A. 8.9 feet/seconds B. 9.8 feet/seconds C. 25.1 feet/second D. 52.1 feet/seconds
goblinko [34]

Answer:

the answer is c) 25.1 feet/second

Explanation:

because the blood droplets can not increase speed past terminal velocity 100.

3 0
2 years ago
Resolve the vector shown below into its components.
yan [13]

Answer:

D. s=3x+4y

Explanation:

The line is at the 3rd column in the 4th row.

7 0
1 year ago
A proton is accelerated down a uniform electric field of 450 N/C. Calculate the acceleration of this proton.
olasank [31]

Answer:

The acceleration of proton will be 431.137\times 10^8m/sec^2

Explanation:

We have given electric field E = 450 N/C

Charge on proton =1.6\times 10^{-19}C

Force on electron due to electric field is given by F=qE=1.6\times 10^{-19}\times 450=720\times 10^{-19}N

Mass of electron m=1.67\times 10^{-27}kg

Now according to second law of motion F=ma

So 720\times10^{-19}=1.67\times 10^{-27}a

a=431.137\times 10^8m/sec^2

So the acceleration of proton will be 431.137\times 10^8m/sec^2

6 0
3 years ago
What best describes an impulse acting on an object?
Triss [41]
The force on an object use Socratic glad to help ..
8 0
2 years ago
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