A column of water exerts a pressure of 100.Pa and a force of 2.40 N. What is the cross-sectional area of a the column?

Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

<em>"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.</em>"

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "</em>

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "</em>

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha

Gwar [14]

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

Work done on the proton =Negative of the change in the electric potential energy of the proton field
In the given case, W = -qΔV
-W = qΔV
= qEcosθ
Therefore, work done on the proton = -e(8.50× $10^2$ N/C)(2.5m)(1)
= -3.40× $10^-^1^6$ J
Any change in the potential energy indicates the work done by the proton.
Therefore the positive sign shows that the potential energy increases when the proton does the work.
The negative sign shows that the potential energy decreases when the proton does the work.

To learn more about electric potential energy, refer

brainly.com/question/14306881

#SPJ4

3 0

1 year ago

To determine power, it is necessary to know the (2 points) a. energy required and the time it takes. b. amount of work and the d

Sveta_85 [38]

the answer A. since power is Joules/seconds and energy is rated in Joules

8 0

3 years ago

Read 2 more answers

Assume the elevator has upward acceleration a, and consider the stack that has two boxes of mass M. What is the magnitude of the

asambeis [7]

Answer:

Explanation:

M(a+g)

4 0

2 years ago

Does the kinetic energy of a car change more when it accelerates from 27 km/h to 37 km/h or when it accelerates from 37 km/h to

scoray [572]

Answer:

3) From 37 km/h to 47 km/h

Explanation:

Kinetic Energy: This is the energy posses by a body in motion. The Unit of kinetic energy is joules. And it is expressed mathematically as,

Ek = 1/2mv².................... Equation 1

ΔEk = 1/2m(v₂² - v₁²)...................... Equation 2

Where ΔEk = change in kinetic energy, m = mass of the body, v₁ = initial velocity, v₂ = final velocity of the body.

<em>Note: The mass of the car is constant in both cases.</em>

When the body accelerate from 27 kh/h to 37 km/h

ΔEk₁ = 1/2m(37²-27²)

ΔEk₁ = 1/2m(640)

ΔEk₁ = 320m J.

When The body accelerate from 37 km/h to 47 km/h

ΔEk₂ = 1/2m(37²-47²)

ΔEk₂ = 1/2m(840)

ΔEk₂ = 420m J

From the above,

ΔEk₂ > ΔEk₁.

Hence the correct option is 3) From 37 km/h to 47 km/h

3 0

3 years ago