PHYSICS COLLEGE

a. A nucleus is made up of protons and neutrons. Protons have positive charges and neutrons have no charge. The strong nuclear force holds the nucleus together because it acts against another force inside the nucleus. What force is the strong nuclear force counteracting?

Answers

Answer 1

Answer:

Answer:

The electromagnetic force tends to push the protons apart. (Like forces repel).

Explanation:

Answer 2

Answer:

Answer:its the electromagnetic force

Explanation:

Protons are positive so repel themselves

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(a) Is the velocity of car A greater than, less than, or the same as thevelocity of car B?
(b) Is the initial position of car A greater than, less than, or equal to the
initial position of car B?
(c) In the time period from t = 0 tot = 1 s, is car A ahead of car B,
behind car B, or at the same position as car B?

Answers

a. ) Is the velocity of car A less than the velocity of car B b. the initial position of car A greater than the initial position of car B c. ahead In the time period from t = 0 tot = 1 s, is car A ahead of car B?.

what is velocity ?

Velocity is the parameter which is different from speed, can be defined as the rate at which the position of the object is changed with respect to time, it is basically speeding the object in a specific direction in a specific rate.

Velocity is a vector quantity which shows both magnitude and direction and The SI unit of velocity is meter per second (ms-1). If there is a change in magnitude or the direction of velocity of a body, then it is said to be accelerating.

Finding the final velocity is simple but few calculations and basic conceptual knowledge are needed.

For more details regarding velocity, visit

brainly.com/question/12109673

#SPJ2

Answer:

a. less than, b. greater than, c. ahead

Explanation:

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=(I)/(\Delta t)=(328.6 kg m/s)/(0.812 s)=404.7 N$

How fast can the car take this curve this curve without skidding to the outside of the curve?

Answers

Lets write the data down. That will help us solve the problem later:

R = 36 m

θ = 18º

m = 1492 kg

μ = 0.67

g = 9.8 m/s²

Lets draw all the forces that act on the car:

In order to the car won't skidding to the outside of the curve we must have the centripetal force equals the friction force:

$F_(cp)=f_a$

$(m.v^(2))/(R)=\mu.F_N$

A shot-putter exerts an unbalanced force of 128 N on a shot giving it an acceleration of 19m/s2. What is the mass of the shot?

Answers

Answer:

128 is the ans cuz N is also lnown as mass

Explanation:

128

A long copper cylindrical shell of inner radius 5 cm and outer radius 8 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 7 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm,
(b) 1.5 cm,
(c) 2.5 cm,
(d) 3.5 cm,
(e) 7 cm.

Answers

Answer:

a. 0

b. 8.4N/C

c. 5.04N/C

d. 3.6 N/C

e. 1.8N/C

Explanation:

The following data are given

inner cylindrical radius,r=5cm

outer cylindrical radius R=8cm

Charge density,p=7pc/m

radius of rod= 1cm

a. at distance 0.5cm from the center of the rod, this point falls on the rod itself and since the charge spread out on the surface of the rod, there wont be any electric field inside the rod itself

Hence E=0 at 0.5cm

b. at 1.5cm i.e 0.015m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.015)\nE=8.4N/C$

The direction of the field depends on the charge on the rod

c. at 2.5cm i.e 0.025m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.025)\nE=5.04N/C$

The direction of the field depends on the charge on the rod

d. at 3.5cm i.e 0.035m this point is still within the rod and the inner cylinder

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.035)\nE=3.6N/C$

The direction of the field depends on the charge on the rod

e. at 7cm which is a point outside the rod and the cylinder, the electric field is

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.07)\nE=1.8N/C$

The direction of the field depends on the charge on the rod

To remove 800j of heat the compressor in the fridge does 500j of work. how much heat is released into the room?

Answers

Answer:

Heat released into the room = 1300 J

Explanation:

CONCEPT:

According to second law of thermodynamics , heat cannot flow from a lower temperature to a higher temperature.But the refrigerator transfers heat from lower to higher temperature .For this , we have to do work on the refrigerator.

This work is used to transfer heat from lower to higher temperature.

Heat removed by fridge = 800 J

Work done = 500 J

Heat released into the room = Heat removed + work done

Heat released into the room = 800 +500

Heat released into the room = 1300 J

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A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from a sodium street lamp, radio waves from an AM radio station, radio waves from an FM radio station, and microwaves from an antenna of a communications system. Rank these types of waves in terms of increasing photon energy, lowest first.

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