A 72-tooth gear is connected to a 12-tooth gear. If the large gear makes one complete turn, how many turns

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

yuradex [85]

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

V=4x8=32V

k.e=eV= 2X32=64eV

3 0

4 years ago

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Total number of oscillation per unit time is called frequency of oscillation. Here, $f=\dfrac{31}{15}=2.06\ Hz$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{15}{4\pi^2\times 2.06^2}$

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

$v_{max}=A\times \omega$

$v_{max}=A\times 2\pi f$

$v_{max}=0.075\times 2\pi \times 2.06$

$v_{max}=0.970\ m/s$

$v_{max}=97\ cm/s$

Hence, this is the required solution.

5 0

3 years ago

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar

Iteru [2.4K]

The direction of the electric field would be south.

qE/m = 115
 E = 115*m/q 
 = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) 
 = 762.87 * 10^(-12) 
 = 6.27 x 10^-10 N/C

Hope this answers the question. Have a nice day. Feel free to ask more questions.

6 0

3 years ago

A 65 kilogram astronaut weighs 638 newtons at the surface of earth what is the mass of the astronaut at the surface of the moon,

Montano1993 [528]

The mass of the astronaut is still 65 kilograms. Mass is constant or doesn't change no matter where you are.

3 0

4 years ago

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