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ELEN
3 months ago
12

Two charges of 15 pC and −40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through

the surface of the cube.
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
4 0

To address this issue, we will utilize the principles related to Gauss' law, which states that the electric flux across a surface corresponds to the object's charge divided by the permittivity of vacuum. In mathematical terms, this can be expressed as

\phi = \frac{Q_{net}}{\epsilon_0}

It's crucial to remember that the net charge equals the difference between the two specified charges, so upon substitution,

\phi = \frac{(15-40)*10^{-12}C}{8.85*10^{-12}C^2/Nm^2}

\phi = 2.82WB

The negative sign indicates that the flux is directed into the surface

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A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts
inna [3103]
Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

3 0
3 months ago
What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?
Keith_Richards [3271]
Based on my findings, within a period of 2 hours, there are certain atoms remaining. N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0 Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0 This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms Consequently, 2.698 * 10^9 atoms represents the value of N0.
4 0
3 months ago
A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air
Sav [3153]

Answer:

Height (h) = 17 m

Velocity (v) = 18.6 m/s

Explanation: This problem can be solved using kinematic motion equations.

Given Data

Initial velocity (u) = 0

Acceleration (a) = g

Time (t) = 1.9 seconds

First, we calculate the height.

s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m

Then, we find the final velocity

v=u+at\\v=0+9.81*1.9\\v=18.639

The acceleration graph is a linear representation described by y=9.8, as it remains constant:

The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):

The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):

These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.

5 0
3 months ago
A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight
kicyunya [3294]

The following values have been provided:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

 

The equation for potential energy is:

E = w h

where h indicates the height that the person needs to ascend, therefore:

h = 3000 / 1067.52

h = 2.81 m

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<span>
</span>

<span>Thus, he must ascend 2.81 meters</span>

3 0
2 months ago
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