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ra1l [238]
3 years ago
7

A strong-armed physics student throws a tennis ball vertically. The ball stays in the air for 5.5 seconds. Assuming the ball lef

t from the ground, how fast did it leave his hand?
Physics
1 answer:
Mila [183]3 years ago
4 0

Answer:

27.5 m/s

Explanation:

applying motion equations we can find the answer,

v = u + a*t

Let assume ,

u = starting speed(velocity)

v = Final speed (velocity)

t =  time taken for the motion

a = acceleration

by the time of reaching the highest point subjected to the gravity , the speed should be equal to zero  (only a vertical speed component is there)

for the complete motion it takes 5.5 s. that means to reach the highest point it will take 5.5/2 =2.75 seconds

we consider the motion upwards , in this case the gravitational  acceleration should be negative in upwards (assume g=10 m/s2)

that is,

v = 0 , a = -10ms^{-2}    , t =2.75

v = u + at

0 = u -10*2.75

u = 27.5 ms^{-1}

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Thomson is responsible for discovering that an atom containing _________
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Two point charges of equal magnitude are 8.0 cm apart. At the midpoint of the line connecting them, their combined electric fiel
bagirrra123 [75]

Answer:

r = 8/2 = 4cm = 0.04m

k = 9×10^9

Enet = 51 N/C

Enet = E1 + E2

since E1 = E2

E1 = Enet/2 = 51/2

E/2 = kq/r²

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Explanation:

The electric field is a region around a

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According to coulomb’s law ,if a unit

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two charges. If the charge q is moved

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7 0
2 years ago
An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
oksian1 [2.3K]

Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A   ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

5 0
3 years ago
Hi guys! :) I really need help on this!! Will def mark brainliest
Travka [436]

Wavelength = (speed) / (frequency)

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3 0
3 years ago
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