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3 years ago

The table lists the mass and charge of a proton and a neutron.

Physics

2 answers:

allsm [11]3 years ago

6 0

Answer:

Explanation:

On Edge

Olegator [25]3 years ago

4 0

The gravitational force is much larger than the electrical force for any distance between the particles.

Explanation:

The gravitational force between two particles is given by:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two particles

r is the separation between the two particles

The electric force between two particles is given by:

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the charges of the two particles

r is the separation between the two particles

In this problem, we are comparing the gravitational force and the electric force between a proton and a neutron.

We must immediately notice that the charge of the neutron is zero, since it is electrically neutral:

q = 0

This means that the electric force between the proton and the neutron is always zero, regardless of their distance. And therefore, since their masses is not zero (and the gravitational force is never zero), the correct answer is

The gravitational force is much larger than the electrical force for any distance between the particles.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

About electrical force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te

Nitella [24]

Answer:

50 N

Explanation:

Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:

The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.

The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

6 0

3 years ago

What term was used to describe the final digit

xxMikexx [17]

Answer:

Significant digits (also called significant figures or “sig figs” for short) indicate the precision of a measurement. A number with more significant digits is more precise. For example, 8.00 cm is more precise than 8.0 cm.

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3 years ago

FREE BRAINLIEST! if you can answer this correctly ill give you brainliest and answer some of the questions you have posted :) th

Temka [501]

Answer:3,600 Newtons

Explanation:

The net force acting on the car is

3×10^3squared

Newtons.

Force is defined as the product of the mass of the body and its aaceleration,⇒F=ma

Substituting the above given values we get,F=(1500kg) (2.0m /s^2 squared)=3000 N=3×10^3 squared N.

N=newtons

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3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

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3 years ago

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