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marshall27 [118]
3 years ago
5

A golfer hits a 42 g ball, which comes down on a tree root and bounces straight up with an initial speed of 15.6 m/s. Determine

the height the ball will rise after the bounce. Show all your work.
Physics
2 answers:
nevsk [136]3 years ago
5 0

Answer:

12.2 m

Explanation:

Given:

v₀ = 15.6 m/s

v = 0 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (15.6 m/s)² + 2 (-10 m/s²) Δy

Δy = 12.2 m

Komok [63]3 years ago
4 0

\LARGE{ \boxed{ \rm{ \green{Answer:}}}}

Given,

  • The initial speed is 15.6 m/s
  • The mass of the ball is 42g = 0.042kg

Finding the initial kinetic energy,

\large{ \boxed{ \rm{K.E. =  \frac{1}{2}m {v}^{2}}}}

⇛ KE = (1/2)mv²

⇛ KE = (1/2)(0.042)(15.6)²

⇛ KE = 5.11 J

|| ⚡By conservation of energy, the potential energy at the highest point will also be 5.11 J, since there is no kinetic energy at the highest point because the ball is not moving (we neglect energy lost due to air resistance, heat, sound, etc.) ⚡||

So, we have:

\large{ \boxed{ \rm{P.E. = mgh}}}

⇛ h = PE/(mg)

⇛ h = 5.11 J /(0.042 × 9.8)

⇛ h = 12.41 m

✏<u>The</u><u> </u><u>ball</u><u> </u><u>will</u><u> </u><u>rise</u><u> </u><u>upto</u><u> </u><u>a</u><u> </u><u>height</u><u> </u><u>of</u><u> </u><u>12.41</u><u> </u><u>m</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

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platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin
posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

a=g=9.8 m/s^2 is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s

For the diver jumping from 10 m, s = 10 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

s=ut+\frac{1}{2}at^2

And since u = 0, it can be reduced to

s=\frac{1}{2}at^2

For the diver jumping from 5 m, s = 5 m, so we find

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For the diver jumping from 10 m, s = 10 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s

5 0
3 years ago
A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

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512212njun][[]

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