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3 years ago

What is electric charge ?

2 answers:

6 0

DescriptionElectric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charge: positive and negative. Like charges repel and unlike attract. An object with an absence of net charge is referred to as neutral.

Nastasia [14]3 years ago

4 0

Answer:

Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric charge is a conserved property; the net charge of an isolated system, the amount of positive charge minus the amount of negative charge, cannot change. Electric charge is carried by subatomic particles. In ordinary matter, negative charge is carried by electrons, and positive charge is carried by the protons in the nuclei of atoms.

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If there is an unbalanced force acting on an object, the object will?

Nonamiya [84]

Answer:

Explanation:

because newton's second law states that if a resultant force acts on an object then, it will accelerate in the direction of the resultant force

4 0

3 years ago

Can someone help me?!!!!!

SOVA2 [1]

Answer:magnitude -5; angle 160°

Explanation:

Vector A is described as having magnitude 5 and angle -20°.

To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.

5 @ -20° = 5 @ 340°

5 @ -20° = -5 @ 160°

The third one is the answer.

8 0

3 years ago

What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency

Anvisha [2.4K]

Answer:

$1.04\times 10^{-7}$ T

Explanation:

$IP$ = Power of the bulb = 100 W

$r$ = distance from the bulb = 2.5 m

$I$ = Intensity of light at the location

Intensity of the light at the location is given as

$I = \frac{P}{4\pi r^{2}}$

$I = \frac{100}{4(3.14) (2.5)^{2}}$

$I$ = 1.28 W/m²

$B_{o}$ = maximum magnetic field

Intensity is given as

$I = \frac{B_{o}^{2}c}{2\mu _{o}}$

$1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}$

$B_{o} = 1.04\times 10^{-7}$ T

7 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago

Consider a situation where the acceleration of an object is always directed perpendicular to its velocity. This means that

Bond [772]

Answer:

this situation would not be physically possible

7 0

3 years ago

What is electric charge ?​

What is electric charge ?