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Inga [223]
3 years ago
12

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten

tial energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 joules. Air friction is negligible. The maximum height reached by the ball is most nearly
Physics
1 answer:
satela [25.4K]3 years ago
6 0

Answer: 20m

Explanation:

We will solve this question by applying the law of conservation of energy which states that the sum of potential energy and kinetic energy is always the same.

The PE is 0 at surface and maximum at top while the KE is maximum at surface and 0 at top.

From the question,

PE = mgh = 50 J -(1)

mg* 10 = 50

mg = 50/10

mg = 5

The total energy at that point = PE + KE = 50 + 50 = 100 J

Therefore, at topmost point, the PE will be 100 J

mgH = 100J , H is the needed height

Using the value of mg obtained above, we have

H= 100/5

H = 20 m

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problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi
Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
3 years ago
What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.0
SOVA2 [1]

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

Energy=h\times frequency

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So,  Formula for energy:

Energy=h\times \frac {c}{\lambda}

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

<u>Wavelength = 736.67 nm</u>

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3 years ago
What kind of model is shown below?
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D. a foot model



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The cylinder has a volume of 37.46 cubic cm
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