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3 years ago

What is the precision for each measurement for the online simulation. What would be the precision for force

Physics

1 answer:

zimovet [89]3 years ago

4 0

Answer:

Unit of precision for force is the Newton.

Explanation:

It is the official unit used to describe force in science and mostly abbreviated with the symbol N.

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How does potential difference behave in a parallel circuit

Illusion [34]

The same voltage will appear across all resistors in parallel.

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3 years ago

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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

nikitadnepr [17]

The radius of curvature of the proton's path while in the field is $66.67$ × $10^{-2}$ .

b) Let R = radius curvature of protons path. Then,

relation b/w B, R, and v is: -

$B = mv/eR\\R=mv/eB$

$R=\frac{1.6*10^{-27} * 20*10^{6}}{1.6*10^{-19}*0.3 }$

$R =66.67$ × $10^{-2}$

Hence, the radius of curvature of the proton's path while in the field is $66.67$ × $10^{-2}$ .

<h3>What do you mean by Magnetic field?</h3>

The magnetic influence on moving electric charges, electric currents and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field. The magnetic field of a permanent magnet pulls on ferromagnetic substances like iron and attracts or repels other magnets. A magnetic field that varies with location will also exert a force on a variety of non-magnetic materials by changing the velocity of those particles' outer electrons. Electric currents, like those utilized in electromagnets, and electric fields that change in time produce magnetic fields that surround magnetized things.

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1 year ago

A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?

faltersainse [42]

Negative

Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.

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2 years ago

When does a scientist create the conclusion?

Lerok [7]

After they have gathered enough information by testing the theory.

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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