PHYSICS COLLEGE

What force causes oppositely charged particles to attract each other? A. Magnetic force B. Compression
C. Electrical Force D. Gravity

Answers

Answer 1
Answer:

Answer:

It is electrical force

Explanation:

i got it wrong on A P E X  with magnetic hope this helps!
Answer 2
Answer:

Answer:

The electromagnetic force.

Related Questions

As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. Since you're an engineer, the first thing you do when you wake up is drill a small hole in the ice and estimate the ice to be 6.7cm thick and the distance to the closest shore to be 30.5 m. The ice is so slippery (i.e. frictionless) that you cannot seem to get yourself moving. You realize that you can use Newton's third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 588 N. (Lucky for you that, as an engineer, you sleep with your knife in your pocket and your boots on.)1)(a) What direction should you throw your boot so that you will most quickly reach the shore? away from the closest shore perpendicular to the closest shore straight up in the air at your friend standing on the closest shore2)(b) If you throw your 1.08-kg boot with an average force of 391 N, and the throw takes 0.576 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)391 N 3)(c) How long does it take you to reach shore, including the short time in which you were throwing the boot?Just number 3
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You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.17 m . Part A Determine the minimum torque that the motor must be able to provide. Express your answer with

Answers

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

We wish to use a motor to lift a 10-kg (m) bund of shingles with an upward acceleration of 1.5 m/s² (a).

The resulting force (F = m.a) is the difference between the tension (T) of the pulley and the weight (w = m.g) of the shingles.

T-m.g=m.a\nT = m.g+m.a = m(g+a) = 10 kg (9.8m/s^(2)+1.5m/s^(2) )=113 N

where,

  • g: gravity

Then, we can calculate the minimum torque (τ) that the motor must apply using the following expression.

\tau = r * T = 0.17m * 113N = 19N.m

where,

  • r: radius of the pulley

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

Learn more: brainly.com/question/19247046

Answer:

\tau=19.21\ N-m

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, a=1.5\ m/s^2

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

T-mg=ma

T=m(g+a)

T=10* (9.8+1.5)

T = 113 N

Let \tau is the minimum torque that the motor must be able to provide. It is given by :

\tau=r* T

\tau=0.17* 113

\tau=19.21\ N-m

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.

Find the current if 20C of charge pass a particular point in a circuit in 10 seconds.

Answers

The current will be "2 A".

Given values are:

  • Charge, Q = 20 C
  • Time, t = 10 seconds

As we know,

→ Current = (Charge)/(Time)

or,

→ i = (Q)/(t)

BY substituting the values, we get

      = (20)/(10)

      = 2 \ A

Thus the answer above is right.

Learn more about current here:

brainly.com/question/19668907

Answer:

2 A

Explanation:

From the question,

Q = it..................... Equation 1

Where Q = Quantity of charge, i = cudrrent, t = time.

Make i the subject of the equation

i = Q/t.......................... Equation 2

Given: Q = 20 C, t = 10 seconds.

Substitute these values into equation equation 2

i = 20/10

i = 2 A.

Hence the current is 2A

A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.20 ?

Answers

Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

Time, t = 1 s

resistance of the coil, R = 1.2 ohm

The induced emf is given by

e=(d\phi)/(dt)=A(B' - B)/(t)

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

i = e/R

i = 2.184/1.2

i = 1.82 mA

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

A 2-C charge experiences a force of 40 N when put at a certain location inspace. The electric field at that location is a. 2 N/C.b. 20 N/C. c. 30 N/C. d.
40 N/C. e. 60 N/C.

Answers

Answer:

E = 20 N/C

Explanation:

Given that,

Charge, q = 2 C

Force experience, F = 40 N

We need to find the electric field at that location.

The electric field in terms of electric force is given by :

F = qE

Where

E is the electric field

E=(F)/(q)\n\nE=(40\ N)/(2\ C)\nE=20\ N/C

So, the electric field at that location is 20 N/C.

What is required for a sound wave to be reflected?

Answers

when sound travels on a certain wave pattern or medium once it hits the surface or surfaces of another wave or mdeium and then it bounces back that is a reflected sound wave.
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