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devlian [24]
3 years ago
6

Which of these forces in carbon-14 isotopes transforms a neutron into a proton? gravitational forces electromagnetic forces weak

nuclear forces strong nuclear forces
Physics
2 answers:
dolphi86 [110]3 years ago
7 0

Answer:

Weak Nuclear Forces

Explanation:

Whenever a neutron gets converted to a proton, electron is also released along with the proton.

n → p + e where n denotes the neutron, p denotes the proton

and e denotes the electron.

This is also called beta decay. This is a weak force since it involves decay.

A strong nuclear force is one which is like the binding of a neutron and the proton in a nucleus.

Allisa [31]3 years ago
5 0
I think the correct answer is the third option. It is the weak nuclear forces in carbon-14 isotopes that transforms a neutron into a proton emitting a beta particle. This weak force is responsible for all radioactive decay and is important in nuclear fission.
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A testable question about what would happen if you fell into a black hole
daser333 [38]

Answer:

You wouldnt fall you would be sucked and you would lose all air supply and you lungs would pop

Explanation:

7 0
2 years ago
Problems - Show all work.
Fittoniya [83]

Answer:

21s

Explanation:

Given parameters;

Radius  = 10m

Speed or velocity  = 3m/s

Unknown:

Period  = ?

Solution:

To solve this problem, use the expression:

      v  = \frac{2\pi r}{T}  

r is the radius

T is the unknown

           Input the parameters and solve for T;

    3  = \frac{2 x \pi  x 10}{T}  

    62.84 = 3T

         T  = 21s

4 0
2 years ago
A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen
Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return  = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

  f_1 = \dfrac{f_0(u + v)}{u}..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

  f_2= \dfrac{f_1(u)}{u-v}

now, inserting the value of equation (1)

  f_2= f_0\dfrac{u+v}{u-v}

  1275=1240\times \dfrac{340+v}{340-v}

  1.02822(340 - v) = 340 + v

   2.02822 v = 340 x 0.028822

   2.02822 v = 9.799

   v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

5 0
3 years ago
A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu
kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the​ rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the​ cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) =  ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i)  work required to pump the contents to the​ rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the​ cone's rim

Extra work required = 28,914.229 lb ×  5ft = 144571.145 lb.ft

Total work required = (281,913.733 +  144571.145) lb.ft

                                 = 426,484.878 lb.ft

5 0
2 years ago
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
3 years ago
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