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Alik [6]
3 years ago
11

Please please please help me

Physics
1 answer:
pogonyaev3 years ago
3 0

Answer:

C) seems to be the correct one

Each alpha particle emitted decreases the mass of the nucleus by 4 and decreases the atomic number by 2.

Also, each beta particle will increase  the atomic number by 1.

So if C emits 5 alpha particles the atomic mass decreases by 20 and the atomic number decreases by 10. Then emission of 2 beta particles brings the atomic number up to minus 8.

So a total of 7 particles have been emitted and the number of alpha particles emitted exceeds the number of beta particles emitted.

You might be interested in
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a
ryzh [129]

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = 0

m₁u₁ = - m₂u₂

u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

3 0
4 years ago
A) Calculate the pressure of water at the bottom of a well if the depth
Molodets [167]

Answer: 98000pa

Explanation:Given,

Depth(h)=10m

gravity(g)=9.8m/s

density(δ)=1000kg/m^3

we know,

P=hdg

P=10*1000*9.8

P=98000pa

7 0
3 years ago
A machine is supplied energy at a rate of 4,000 W and does useful work at a rate of 3,760 W. What is the efficiency of the machi
Dmitry_Shevchenko [17]
                = (3,760 joule/sec) / (4,000 joule/sec)
 
               =    3,760 / 4,000  =  0.94  =  94%
8 0
4 years ago
Which statement correctly describes variable stars?
Kryger [21]

Answer:

D)

Explanation:

The Period-Luminosity relationship tells us that luminosity increases with the period, and of course the more luminosity a star has the more far away they can be seen, so from this we know that:

A) False since lower luminosities can be observed when they are close.

B) False since longer periods means higher luminosities

C) False since lower luminosities can be observed when they are close.

D) True: Variable stars with shorter periods have lower luminosities, so they can only be observed when they are close.

5 0
3 years ago
PLEASE HELP
Gennadij [26K]

Answer:

3.675 m

Explanation:

a_{x} =0 v_{xo}=100 a_{y} =-g  v_{yo}=0

X-direction     | Y-direction

R=x_{o}+ v_{xo} t  | y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2

75=100t         |y=0+0+\frac{1}{2} (9.8)(0.75)

\frac{75}{100} =t             | y=3.675 m

0.75s=t              

Hope it helps

3 0
3 years ago
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