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1 year ago

3. Saagar gives a certain teacher some sass followed by an eye roll. Suddenly, a shoe is thrown at him.

Physics

1 answer:

joja [24]1 year ago

8 0

The average acceleration of the shoes thrown at him, a =2.775 m/s².

<h3>Equation :</h3>

To find the acceleration with given data we use this formula,

v = u + aΔt

Where,

v is final velocity

u is initial velocity

a is acceleration

Δt is time

So,

v - u = aΔt

a = v - u /Δt

a = 12km/hr - 0 / 1.2 s

a = 12km/hr / 1.2s

Changing km/hr into m/s,

1 kilometer/hour = 1000meters / 3600 seconds

We get,

12km/hr = 3.33m/s

a = 3.33 m/s / 1.2s

a = 2.775 m/s²

<h3>What is velocity ?</h3>

When an object is moving, its velocity is the rate at which its direction is changing as seen from a specific point of view and as measured by a specific unit of time. Velocity is the rate and direction of an object's movement, whereas speed is the time rate at which an object is moving along a path.

To know more about speed :

brainly.com/question/28224010

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A car starts from rest and accelerates to 14 m/s in 2 seconds. What was its acceleration

Romashka [77]

Answer:

7m/s^2

Explanation:

using v=u+at

since the car started from rest, u=0 , v=14m/s t=2s

a =acceleration.

14=0+a×2

14=0+2a

14=2a

a= 14/2 =7

a=7m/s^2

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2 years ago

A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of

grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

$KE = \frac{1}{2} mv^2$

Replacing with our values we have that

$KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2$

$KE = 1.1935*10^{-13}J$

Therefore the kinetic energy of the helium nucleus is $1.1935*10^{-13}J$

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

$KE = qV$

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

$V = \frac{KE}{q}$

$V = \frac{1.19*10^{-13}}{1.6*10^{-19}}$

$V = 743750eV$

Therefore the kinetic energy in electron vols is $743750eV$

PART C) Applying the same relationship but now using the Helium core load, we will have to

$KE = QV$

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

$V = \frac{KE}{Q}$

But we need to note that the charge is equal to the number of charge for the unit charge, then

$Q = \text{No. Charge} \times \text{Unit Charge}$

$Q = (2)(1.6*10^{-19}C)$

$Q = 3.2*10^{-19}C$

Now replacing we have that

$V= \frac{1.19*10^{-13}}{32*10^{-19}}$

$V = 371875V$

Therefore the voltage applied is 371875V

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Explanation:

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Answer:

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Explanation:

I hope it's helpful!

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