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ella
1 month ago
12

Assume that charge −q is placed on the top plate, and +q is placed on the bottom plate. What is the magnitude of the electric fi

eld E between the plates? Express E in terms of q and other quantities given in the introduction, in addition to ϵ0 and any other constants needed..
Physics
1 answer:
Ostrovityanka [3.2K]1 month ago
8 0
Let A represent the area of each plate. According to Gauss's Law, the electric field present between the plates can be derived.
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A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk
kicyunya [3294]

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

r = 16m

m = 82kg

\mu_s = 0.63

The maximum velocity can be determined using centripetal force,

F_c = \frac{mv^2}{r}

Should be equal to,

\frac{mv^2}{r} = \mu_s mg

v = \sqrt{\mu_s gr}

v = \sqrt{(0.63)(9.8)(16)}

v = 9.93m/s

As a result, the highest speed achievable through the arc without slipping is 9.93m/s

3 0
2 months ago
Six pendulums of mass m and length L as shown are released from rest at the same angle (theta) from vertical. Rank the pendulums
Sav [3153]

Answer:   1m, 1m, 2m, 2m, 4m, 4m.

It’s important to remember that the masses attached do not influence the number of oscillations.

Explanation:

To determine the number of oscillations (complete cycles), we can apply the formula n = t / T ……equation 1

The variables that impact the period of a simple pendulum are solely its length and gravitational acceleration. The period remains unaffected by factors such as mass.

period (T)= 2 x π x √(L/g) ….equation 2

where π = 3.142, L= rope length, and g = 9.8 m/s (gravitational acceleration)

According to the question, the time (t) is 60 seconds.

By merging equations 1 and 2, we obtain  

number of oscillations = time / (2 x π x √(L/g))

Case 1: for L = 4m

number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))

= 14.9 = 14 complete cycles (the problem specifies complete cycles)

Case 2: for L = 2m

number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))

= 21.4 = 21 complete cycles

Case 3: where L = 4m, results in the same as case 1, yielding 14 complete cycles

Case 4: where L = 2m, mirrors the outcome in case 2, producing 21 complete cycles

Case 5: in the instance of L = 1m

number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))

= 30.1 = 30 complete cycles

Case 6: when L = 1m, which repeats case 5, also gives 30 complete cycles

From these findings, the order of the pendulums from the highest to lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.

Remember, the number of oscillations is independent of their respective masses.

3 0
2 months ago
Read 2 more answers
Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his
Keith_Richards [3271]
Velocity = 71 meters per minute (MPM)
S stands for Speed
D means Distance
T represents Time
To calculate Speed, divide Distance by Time.
6 0
2 months ago
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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
serg [3582]

Response:

The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is \sqrt{725} ft/ sec

Clarification:

Provided:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec. 5 = 125 ft

x(5) = 10 ft/sec. 5 = 50 ft

At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Now, let's calculate the derivative of distance in relation to time

\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

By plugging in the values for h(t) and x(t) and simplifying, we arrive at,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t} = \frac{725}{\sqrt{725}} = \sqrt{725} ft / sec

5 0
1 month ago
An electron is projected with an initial speed of 3.9 × 105 m/s directly toward a proton that is fixed in place. If the electron
Keith_Richards [3271]

Explanation:

Attached is a document that contains the solution.

8 0
1 month ago
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