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nirvana33 [79]
2 years ago
7

A young diver is practicing his skills before an important team competition. Use the diagram below in order to analyze the energ

ies of the diver and complete the statements below.
Where m = mass (kg), g = 9.8 m/s2, v = velocity (m/s), h = height (m), KE = kinetic energy (J), and GPE = gravitational potential energy (J).
Use the equations above to answer the following questions.

A diver with a mass of 90 kg is at a height of 10 m, and he has not jumped off of the board yet (v = 0 m/s). When the diver reaches a height of 5 m (Point C), his gravitational potential energy is

A. 1350 J

B. 8820 J

C. 4410 J

D. 0 J

and his velocity is

E. 4.5 m/s

F. 0 m/s

G. 3.2 m/s

H. 9.9 m/s

Please help will mark brainliest

Physics
2 answers:
Butoxors [25]2 years ago
6 0
The answer of a young diver is practicing his skills before an important team competition use the diagram below in order to analyze thee energies of the D
xxTIMURxx [149]2 years ago
6 0

Answer: The potential energy is C 4410 J and the Velocity is 4.5 m/s

Explanation:

I hope this helps:)

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How has the farming industry been affected by the use of fossil fuels?
Mashutka [201]

Fossil fuels have led to an increase in food costs.

Explanation:

Buried organic material over million of years form combustible geologic deposits of organic materials called fossil fuels that can be converted crude oil, natural gas, or coal.

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3 0
3 years ago
I need help this is for economics, can anyone help me ?
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i can <u>when do u need it by</u> working on it now!!!

4 0
3 years ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla
lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball u=3.2\ m/s

Plane is inclined at an angle 20^{\circ}

To win the Game we need to hit the target at x=2.4\ m away

Launch angle of ball \theta

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

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Range of Projectile is given by

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for R=2.4\ m

2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}

\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}

\sin 2\theta =0.7855

2\theta =51.77

\theta =25.88^{\circ}

so ball must be launched at an angle of 25.88^{\circ}

4 0
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