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3 years ago

Which types of forces exist between the two protons in a helium nucleus?

Physics

1 answer:

777dan777 [17]3 years ago

5 0

Answer:

Option (2) is correct.

Explanation:

When two charges have equal charges they repel each other. On the other hand, if they have unequal charges they attract each other.

The force between the two protons in a helium nucleus is electrostatic charges and the force is repulsive in nature.

Also, there exists a strong nuclear force between charges.

So, a repulsive electrostatic force and an attractive strong nuclear force is the correct answer.

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The freezing point of water is a (physical/chemical) property. The ability of oxygen to react with iron to cause rust is a (phys

AleksandrR [38]

So the freezing point of water is physical and the ability of oxygen to react with iron to cause rust is chemical

7 0

3 years ago

Read 2 more answers

A 2 kg ball travelling West at 6 m/s collides with a 3 kg ball travelling South at 4 m/s stick together and move off with a comm

Alchen [17]

Answer:

Momentum of 2 kg ball:

velocity = 6 m/s

momentum (p)= mv

p = (2)(6) kg m/s

p = 12 kg m/s

kg m/s can also be written as 'N'

Force of 2 kg ball = 12 N

Momentum of 3 kg ball:

velocity = 4 m/s

momentum (p) = mv

p = (3)(4)

p = 12 kg m/s

Since kg m/s can also be written as 'N'

Force of 3 kg ball = 12 N

Now that we have the force applied by both the balls, we can find the resultant force using vector addition

2 kg ball's vector = -12 i

3 kg ball's vector = -12 j

Adding both the vectors, we get:

Resultant vector = -12 i -12 j

The speed both the balls will move at, is the magnitude of the resultant vector

Magnitude of the resultant vector:

|R|² = (i vector)² + (j vector)²

|R|² = (-12)² + (-12)²

|R|² = 144 + 144

|R|² = 288

|R| = √288

|R| = 17 m/s (approx)

The balls will move at a velocity of 17 m/s

Direction of the Common velocity:

TanΘ = Opposite / Adjacent

TanΘ = 12 / 12

Tan Θ = 1

Θ = Arctan(1)

Θ = 45 degrees

Therefore, the common velocity will be 45 degrees down from the horizontal

5 0

4 years ago

10 example of motion with reference

tankabanditka [31]

Answer:

1.Runnng

2.Writing

3.Swimming

4.eating

5.Drinking

6.Moving cars

7.Playing

8.Throwing ball

9.TYping

10.Cycling

Explanation:

7 0

3 years ago

A 80kg astronaut is training in human centrifuge to prepare for a launch. The astronaut uses the centrifuge to practice having a

inessss [21]

The answers on the model of the human centrifuge ready for the launch to each question of the statement are listed below:

a) A force of 2479.210 newtons is acting on the astronaut's back.

b) A net centripetal force of 2479.210 newtons is acting on the astronaut.

c) The centripetal acceleration of the astronaut is 30.990 meters per square second.

d) The astronaut has a linear speed of approximately 19.284 meters per second.

e) The angular speed of the astronaut is 1.607 radians per second (15.346 revolutions per minute).

<h3>How to apply Newton's laws to analyze a process in a human centrifuge training</h3>

The human centrifuge experiments a centripetal acceleration when it reaches a peak angular speed. In this question we must apply Newton's laws of motion and concepts of centripete and centrifugal forces to answer the questions. Now we proceed to answer the questions:

<h3>How much force is acting on the astronaut's back?</h3>

By the third Newton's law the astronaut experiments a rection force (F), in newtons, which has the same magnitude to centrifugal force but opposed to that force. The magnitude of the force acting on the back of the astronaut is equal to:

$F = 3.16\cdot (80\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2479.210\,N$

A force of 2479.210 newtons is acting on the astronaut's back. $\blacksquare$

<h3>What is the net centripetal force on the astronaut?</h3>

By the second and third Newton's laws we know that the net centripetal force on the astronaut is equal to the magnitude of the force found in the previous question. Thus, a net centripetal force of 2479.210 newtons is acting on the astronaut. $\blacksquare$

<h3>What is the astronaut's centripetal acceleration?</h3>

The centripetal acceleration of the astronaut (a), in meters per square second, is found by dividing the result of the previous question by the mass of the astronaut (m), in kilograms:

$a = \frac{F}{m}$ (1)

If we know that F = 2479.210 newtons and m = 80 kilograms, then the centripetal acceleration of the astronaut is:

$a = \frac{2479.210\,N}{80\,kg}$

$a = 30.990\,\frac{m}{s^{2}}$

The centripetal acceleration of the astronaut is 30.990 meters per square second. $\blacksquare$

<h3>What is the astronaut's linear speed?</h3>

By definition of uniform circular motion, we have the following formula for the linear velocity of the astronaut (v):

$v = \sqrt{a\cdot r}$ (1)

Where r is the radius of the human centrifuge, in meters.

If we know that $a = 30.990\,\frac{m}{s^{2}}$ and $r = 12\,m$ , then linear velocity of the astronaut is:

$v = \sqrt{\left(30.990\,\frac{m}{s^{2}} \right)\cdot (12\,m)}$

v ≈ 19.284 m/s

The astronaut has a linear speed of approximately 19.284 meters per second. $\blacksquare$

<h3>What is the astronaut's angular speed? </h3>

The angular speed of the astronaut (ω), in radians per second, is found by the following kinematic relationship:

$\omega = \frac{v}{R}$ (1)

If we know that v ≈ 19.284 m/s and R = 12 m, then the angular speed is:

$\omega = \frac{19.284\,\frac{m}{s} }{12\,m}$

ω = 1.607 rad/s (15.346 rev/m)

The angular speed of the astronaut is 1.607 radians per second (15.346 revolutions per minute). $\blacksquare$

To learn more on centripetal forces, we kindly invite to check this verified question: brainly.com/question/11324711

6 0

2 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2

Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0

4 years ago