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4 years ago

Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal. What is the range of the ball?

Physics

2 answers:

musickatia [10]4 years ago

6 0

4.8miliseconds broooo.

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Extra info related to the question :-

What is a rationlising factor ?

If the product of two irrational numbers is a rational number then each of the the two is rationalising factor of the other.

What is the purpose of rationilising the denominator ?

If the denominator of a fraction is irrational it is somewhat difficult to locate the fraction on number line. So, to make it easy we need to rationalise the denominator.

How to rationalise the denominator ?

1) Consider the denominator of the fraction.

2) Then multiply the numerator and denominator with rationalising factor of the denominator.

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just olya [345]4 years ago

4 0

Answer:

the range or the ball is 48.81 m

Explanation:

given;

Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.

find:

What is the range of the ball?

solution:

let Ф = 25°

Vo = 25 m/s

consider x-motion using time of fight: x = Vox * t

where x = R = range

t = 2 Voy

R = Vo² sin (2Ф)

plugin values into the formula:

R = (25)² sin (2*25)

9.81

R = 48.81 m

therefore, the range or the ball is 48.81 m

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A gamma-ray photon produces an electron and a positron, each with a kinetic energy of 261 keV . h=6.626×10−34J⋅s, c=2.998×108m/s

ddd [48]

Answer:

The energy and the wavelength of the photon are 1.546 MeV and $8.036\times10^{-13}\ m$ .

Explanation:

Given that,

Kinetic energy = 261 KeV

Planck's constant $h = 6.626\times10^{−34}\ J.s$

Speed of light $c=2.998\times10^{8}\ m/s$

Mass of electron $m_{e}=9.109\times10^{-31}\ kg$

Charge $q=1.602\times10^{-19}\ C$

(A). We need to calculate the energy of the photon

Using formula of rest mass energy

$E=m_{0}c^2$

$E=9.109\times10^{-31}\times(3\times10^{8})^2$

$E=8.198\times10^{-14}\ J$

Energy in eV

$E=\dfrac{8.198\times10^{-14}}{1.6\times10^{-19}}$

$E=512375\ eV$

$E=0.512\ MeV$

The total energy of photon

$TE=2(E+K.E)$

$TE=2(0.512+0.261)$

$TE=1.546\ MeV$

(B). We need to calculate the wavelength of the photon

Using formula of wavelength

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.626\times10^{−34}\times3\times10^{8}}{1.546\times10^{6}\times1.6\times10^{-19}}$

$\lambda=8.036\times10^{-13}\ m$

Hence, The energy and the wavelength of the photon are 1.546 MeV and $8.036\times10^{-13}\ m$ .

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