A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Answer:

σ₁ = C/m²

σ₂ =  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

Charge (Q₁) = 572.8 C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴

  = C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = = =  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ = = = C/m²

K=E[(m+M)/M] Kmin=4.4

To tackle this question, we know the following:

1 Albert equals 88 meters.

1 A = 88 m.

Initially, we square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Thus, one acre is approximately 0.52 square alberts.</span>

Answer:

option D.

Explanation:

The correct choice is option D.

For an object in equilibrium, the torque measured at any point will be zero.

An object is deemed to be in equilibrium when the net moment acting on it equals zero.

If the object experiences a net moment not equal to zero, it will rotate and will not remain stable.

a)

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top

= loop radius

By solving for v,

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

Given the plane's deceleration is consistent, we can obtain it using the following equation:

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

- The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B:

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B: