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stepan [7]
3 years ago
15

A cannonball is fired horizontally from the top of a cliff. the cannon is at height h = 80.0 m above ground level, and the ball

is fired with initial horizontal speed v0. assume acceleration due to gravity to be g = 9.80 m/s2 .assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. what is the y position of the cannonball at the time tg/2?
Physics
1 answer:
SOVA2 [1]3 years ago
6 0

Answer;

= 60 m

Explanation;

s = 80 m

= ½at² = ½ * 9.8m/s² * t² → t = 4.04 s  

At any time 0 ≤ t ≤ 4.04 s, height is  

y = Yo + ½at² = 80 m - 4.9m/s² * t²  

so at t = 4.04s/2 = 2.02 s,  

y = 80 m - 4.9m/s² * (2.02s)² = 60 m  

(so it drops 15 m during the first 2.02 s and 60 m during the second 2.02 s)

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Doing work' is a way of transferring energy from one object to another, energy is transferred when a force moves through a distance.

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An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase
goldenfox [79]

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changing the magnetic field more rapidly

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8 0
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discharge occurs when oppositely charged objects get close enough for the air between them to become electrically charged.
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4 0
3 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
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