A parallel-plate capacitor is constructed using adielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00X108V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.
Answer:
A = 0.188 m²
Explanation:
First we find the distance between the plates by using the formula of electric field intensity:
E = ΔV/d
d = ΔV/E
where,
d = distance between plates = ?
ΔV = Potential Difference = 4 KV = 4000 V
E = Electric Field = 2 x 10⁸ V/m
Therefore,
d = 4000 V/(2 x 10⁸ V/m)
d = 2 x 10⁻⁵ m
Now, we find the Area of Plates by using formula of capacitance:
C = A∈₀∈r/d
where,
C = Capacitance = 0.25 μF = 0.25 x 10⁻⁶ F
A = Area of Plates = ?
∈₀ = Permittivity of free space = 8.85 x 10⁻¹² C/N.m²
∈r = Dielectric Constant = 3
Therefore,
0.25 x 10⁻⁶ F = A(8.85 x 10⁻¹² C/N.m²)(3)/(2 x 10⁻⁵ m)
A = (0.25 x 10⁻⁶ F)(2 x 10⁻⁵ m)/(3)(8.85 x 10⁻¹² C/N.m²)
A = 0.188 m²
Which of the following equations accurately defines acceleration?
Plz help ASAP
Answer:
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Explanation:
Answer: i think its b
sorry if im wrong
Explanation:
When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the average force was 72.9 N, how much time was it in contact with the floor?
(Unit = s)
Remember: up is +, down is -
Answer:
t = 0.0689 s
Explanation:
Given that,
Mass of a basketball, m = 0.622 kg
Initial velocity, u = 4.23 m/s (downward or negative)
Final velocity, v = 3.85 m/s (up of positive)
Average force, F = 72.9 N
We need to find the time it was in contact with the floor. The force is given by :
[tex]F=ma\\\\F=m\dfrac{v-u}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-(-4.23))}{72.9}\\\\t=0.0689\ s[/tex]
So, the time of contact is 0.0689 s.
a block weighing (Fg) 50 N is resting on a steel table (us = 0.74). The minimum force to start this block moving is what N
Answer:37
Explanation:
Colette launches an air rocket in the upward, positive direction. It launches
with an initial velocity of 25.5 m/s. It accelerates in the downward, negative
direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of
the rocket's displacement?
A) 29 meters
B) 31 meters
C) –150 meters
D) 150 meters
Answer:
b
Explanation:
Answer:
The answer is A) 29 meters
Explanation:
I got this question right on the test! :)
Which of the following charts correctly compares plant and animal cells?
Answer:
Wheres the charts??
Explanation:
Whet net force is required to accelerate a car at a rate of 10 m/s2 if the car
has a mass of 5,000 kg?
Answer:
[tex]\boxed {\boxed {\sf 50,000 \ Newtons }}[/tex]
Explanation:
Force can be found by multiplying the mass by the acceleration.
[tex]F=m*a[/tex]
The mass of the car is 5,000 kilograms and it's acceleration is 10 meters per square second.
[tex]m= 5,000 \ kg \\a= 10 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 5,000 \ kg * 10 \ m/s^2[/tex]
Multiply.
[tex]F= 50,000 \ kg*m/s^2[/tex]
1 kilogram meter per square second is equal to 1 Newton. So, our answer of 50,000 kg*m/s² is equal to 50,000 Newtons[tex]F= 50,000 \ N[/tex]
A net force of 50,000 Newtons is required to accelerate a 5,000 kilogram car at 10 meters per square second.
A 0.050 kg ball starts from rest at some unknown height on a toy roller coaster.
At a later time, it travels through the top of a loop at 2 m/s and a height of 0.40 m.
Since this track is frictionless, what was the starting height of the ball?
Answer:
The starting height of the ball is approximately 0.604 m
Explanation:
The given parameters are;
The mass of the the ball = 0.050
The speed with which it travels through the top loop = 2 m/s
The given height at which the ball moves at 2 m/s = 0.40 m
Therefore, we have;
1/2·m·v² = m·g·h
1/2·v² = g·h
h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204
The additional height = h = 0.204 m
Therefore;
The starting height of the ball ≈ The given height at which the ball moves at 2 m/s + h
The starting height of the ball ≈ 0.40 + 0.204 = 0.604 m
The starting height of the ball ≈ 0.604 m.
When any object in rest, then potential energy is present. But when object is in motion then object have kinetic energy.
Starting height of the ball is 0.604 m.
We know that, when any object is start from rest, then potential energy is converted into kinetic energy.
[tex]\frac{1}{2}mv^{2} =mgh[/tex]
Where m is mass of object, g is gravitational acceleration , h is height and v is velocity of object. (value of g = 9.81 m/ second square)
from above equation,
we get, extra height [tex]h=\frac{v^{2} }{2g} \\\\h=\frac{4}{2*9.81}\\\\h=0.204[/tex] meter
The starting height of the ball will be sum of the height at which ball moves 2 m/s and extra height.
Starting height = 0.40 + 0.204 = 0.604 meter.
Learn more:
https://brainly.com/question/18963960
(8th grade HELP)
1. Inertia causes a stationary object to...
a.stay still
b.move
c.have an increased velocity
d.change it’s speed or direction
2. Once an applied force causes an object to start moving, the object keeps moving because....
a.none of the above
b.the force continues to be applied to it
c.no other force is acting on it
d.it has inertia
Calculate the average speed of a runner who runs to for 500 meters in 40 second
Answer:
12.5
Explanation:
the specific heat of gold is 0.031 calories degrees Celsius and the specific heat of silver is 0.057 calories degrees Celsius so if equal. amounts of each metal are exposed to equal heating which will heat up faster
Answer:
Explanation:
Given:
Specific heat of gold = 0.031cal/°C
Specific heat of silver = 0.057cal/°C
To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.
It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.
Now,
The higher the specific heat capacity the more energy it is required to heat up the substance.
So, Gold with a specific heat capacity of 0.031cal/°C will heat up faster.
PLEASE HELP ME IM TIMED
How much power will it take to move a 10 kg mass at an acceleration of 2 m/s² a distance of 10 meters in 5 seconds?
Answer:
100 Watts
Explanation:
These equations are needed to work out the answer:
power= work done/ time takenwork done= force* distanceforce= mass* accelerationforce: 10 kg* 2m/s= 20
work done: 20* 10m=200
power: 200/2=100
The density of an object is dependent upon the object’s mass and ---
Answer:Volume
Explanation:
Density = mass/ Volume
Answer:
Volume
Explanation:
A fish is 4.7 cm from the front surface of a fish bowl of radius 21 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl? Do not forget the proper sign. (Give your answer in cm.)
______ cm
Where does the fish appear to be when it is 38.9 cm from the front surface of the bowl? (Give your answer in cm.)
______ cm
Answer:
Explanation:
From the information given:
We can properly determine the distance where the fish appear in the air viewing it from in front of the bowl by using the formula:
[tex]\dfrac{n_i}{d_o}+\dfrac{n_2}{d_1}= \dfrac{n_2-n_1}{r}[/tex]
where;
[tex]n_1[/tex] = refractive index in the air; = 1.33 &
[tex]n_2[/tex] = refractive index in water. = 1
[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]
[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{4.7\ cm}[/tex]
[tex]\dfrac{1}{d_i}= - 0.26726 \ cm[/tex]
[tex]d_i =\dfrac{1}{ - 0.26726 \ cm}[/tex]
[tex]\mathbf{d_i }[/tex] = - 3.74 cm
2)
To determine where the fish appear to be when it is 38.9 cm from the front surface of the bowl by using the formula:
[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]
[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{38.9\ cm}[/tex]
[tex]\dfrac{1}{d_i}=- 0.0184759 \ cm[/tex]
[tex]d_i = \dfrac{1}{- 0.0184759 \ cm}[/tex]
[tex]\mathbf{d_i = }[/tex] -54.12 cm
6th grade science I mark as brainliest.
Answer:
Solution:-Distance =400m
Time=20s
We need to find speedAs we know that
[tex]{\boxed{\sf speed \dfrac {Distance {}_{(d)}}{Time {}_{(t)}}}}[/tex]
Substitute the values[tex]\LARGE\leadsto\sf speed=\dfrac {400}{20}[/tex]
[tex]\LARGE\leadsto\sf 20m/s[/tex]
A chemical bond resulting from the exchange of electrons between two atoms is a(n) ________________ bond.
a
covalent
b
ionic
c
Lewis
d
metallic
Answer:
a covalent
Explanation:
ionic bonding involves only the transfer of electrons but not exchange of electrons
plz help me with question ;-;
Answer:
C
Explanation:
A car travels 3500 m in 200 seconds what is the car speed
Answer:
17.5 m/s
Explanation:
We can calculate the meters per second by dividing the distance by time. 3500 divided by 200 is 17.5, therefore the speed is 17.5 meters per second.
a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.
Answer:
Average speed = ( 2V + V1 + V2)/4
Explanation:
Given that a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.
Since the distance is covered at equal intervals of time, and
Speed = distance/time
For the first half distance,
V = distance/t
Cross multiply
Distance = Vt
For the second half distance
(V1 + V2)/2 = distance/t
Distance = t(V1 + V2)/2
The average speed = total distance/ total time.
Average speed = [Vt + t( V1 + V2)/2] ÷ 2t
Average speed = (2Vt + V1t + V2t)/4t
Average speed = t( 2V + V1 + V2)/4t
Time t will cancel out
Average speed = ( 2V + V1 + V2)/4
ANSWER QUICK!!
describe two uses for microwave radiation
heat food, warm water
Atoms of two different elements must have different
A. Electrical charges
B. Number of neutrons
C. Atomic numbers
D. Energy levels
Explanation:
C. Atomic numbers....
Answer:
atomic numbers
Explanation:
In order for two atoms to be different, they have to have a different number of protons. Protons are represented by the atomic number. Thus, atoms of two DIFFERENT elements must have different atomic numbers.
I took the test and got 100%
Hope this helps!
what is the mass and volume of 1000kg/m3 of water?
Answer: The mass would be 1000m3 and the volume would be 1000kg
Explanation:
PLEASE HELP ME IM TIMED
Answer:
the answer is the core
Explanation:
the core is composed of iron and nickel
6th grade science I mark as brainliest.
Answer:
2m 13[tex]\frac{1}{3}[/tex]s
Explanation:
1.5m = 1s
200m = [tex]\frac{200}{1.5}[/tex] × 1s
= 133[tex]\frac{1}{3}[/tex]s
= 2m 13[tex]\frac{1}{3}[/tex]s
Question 12 of 20
How does decreasing the length of a wire affect a circuit?
A. It reduces the resistance caused by the wire.
B. It reduces the voltage carried by the wire.
O C. It increases the resistance caused by the wire.
D. It increases the current passing through the wire.
If we decrease the length of the wire in a circuit, It reduces the resistance.
What is resistance?Resistance is the opposition offered to the flow of current by a circuit element. The factors that affect resistance include;
Length of the wireCross sectional area of the wireHence, if we decrease the length of the wire in a circuit, It reduces the resistance caused by the wire since the resistance depends on the length of the wire.
Learn more about resistance: https://brainly.com/question/15067823
Answer: A
Explanation:
How much Tension force is required to pull a 1500 kg car (it is being towed) forward with an acceleration of 3 m/s^2 if the friction force on the towed car's tires is pulling backward with a force of 2500 N?
If the pull is done horizontally, then the net force on the car is
∑ F = T - f = (1500 kg) (3 m/s²)
where T is the magnitude of the tension in the towing cable, and f is the friction which points in the opposite direction. Then
T = f + (1500 kg) (3 m/s²)
T = 2500 N + 4500 N
T = 7000 N
How do u know that liquid exerts pressure ?
Answer:
All fluids exert pressure like the air inside a tire. The particles of fluids are constantly moving in all directions at random. As the particles move, they keep bumping into each other and into anything else in their path. These collisions cause pressure, and the pressure is exerted equally in all directions.
Explanation:
Hope this helps!! Please consider marking brainliest! Have a good one!!State Newton's three laws of motion in your own words and give an example for each one that helps to explain it in everyday life.
Answer:
The first law is an object won’t change its place unless a force hits it
The second law is the force on an object equivalent to the mass times its acceleration
The third law is when two object meet, they both apply forces on each other making them go in opposite directions
Explanation:
A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?
Answer:
t = 9.14 s
Explanation:
We first analyze the accelerating motion by applying first equation of motion:
Vf₁ = Vi₁ + a₁t₁
where,
Vf₁ = Final Speed of Car before turning off engine
Vi₁ = Initial Speed of Car = 0 m/s
a₁ = acceleration of car = 25 m/s²
t₁ = time taken in accelerating motion
Therefore,
Vf₁ = 25t₁ ---------- equation (1)
Now, we apply second equation of motion:
s₁ = Vi₁ t₁ + (1/2)a₁t₁²
where,
s₁ = distance covered during accelerating motion
Therefore,
s₁ = (0)t₁ + (1/2)(25)t₁²
s₁ = 12.5 t₁² ----------- equation (2)
Now, we analyze the decelerating motion by applying first equation of motion:
Vf₂ = Vi₂ + a₂t₂
where,
Vf₂ = Final Speed of Car = 0 m/s
Vi₂ = Initial Speed of Car after turning off engine
a₂ = deceleration of car = - 3 m/s²
t₂ = time taken in decelerating motion
Therefore,
Vi₂ = 3t₂ ---------- equation (3)
Now, we apply second equation of motion:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₂ = distance covered during decelerating motion
Therefore,
s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²
s₂ = Vi₂ t₂ - 1.5 t₂²
using equation (3):
s₂ = 3 t₂² - 1.5 t₂²
s₂ = 1.5 t₂² ------------ equation (4)
Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):
Vf₁ = Vi₂
using equation (1) and equation (3):
25 t₁ = 3 t₂
t₁ = 0.12 t₂ ------------ equation (5)
Also, we know that sum of the distances is 200 m:
s₁ + s₂ = 200
using equation (2) and equation (4):
12.5 t₁² + 1.5 t₂² = 200
using equation (5):
12.5 (0.12 t₂²) + 1.5 t₂² = 200
3 t₂² = 200
t₂² = 200/3
t₂ = 8.16 s
substitute this in equation (5):
t₁ = 0.12(8.16 s)
t₁ = 0.97 s
Hence, the minimum time required for this motion is:
t = t₁ + t₂ = 0.97 s + 8.16 s
t = 9.14 s
