Answers
Answer: C = 1.319×10^-11 F
Explanation: The formulae that relates the capacitance of a capacitor to the area of the plates, distance between the plates and dielectric constant is given as
C = kε0A/d
Where C = capacitance of plates =?
k = dielectric constant = 2.0
Area of plates = 16.4cm² = 0.00164 m²
d = distance between plates = 2.2 mm = 0.0022m
By substituting the parameters, we have that
C = 2 × 8.85×10 ^-12 ×0.00164/ 0.0022
C = 0.029028 × 10^-12/ 0.0022
C = 13.19× 10^-12
C = 1.319×10^-11 F
Related Questions
If a pressure gauge measure an increase in 3×10^(5)Pa on an area of 0.7 m^2, calculate the increase in the force applied to the area?
What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *1 point45 N450 N450 kg10N
The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)
What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec
A.0.41 sec
B.41 sec
C.4.1 sec
D.4 sec
Answers
A horizontal baseball pitch is launched at 44 m/s. The ball will stay for 4.1 sec (approx) in the air. Hence, option C is correct.
What is Velocity?
The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.
Its SI unit is represented as m/s, and it is a vector quantity, it means that it has both magnitude and direction.
According to the question, the given values are :
Initial Velocity, u = 44 m/s,
Distance travelled, s = 18 m and,
Final velocity, v = 0.
Use equation of motion :
v = u + at
0 = 44 + (-9.8)t
t = 44 / 9.8
t = 4.3 (approx)
Hence, the time for which the ball stay in the air is 4.1 sec (approx).
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Answer:
a 0.41
plug number into equation
Answers
Answer:
A. 29.7 m/s
B. 6.06 s
Explanation:
From the question given above, the following data were obtained:
Maximum height (h) = 45 m
A. Determination of the initial velocity (u)
Maximum height (h) = 45 m
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) = 0 m/s (at maximum height)
Initial velocity (u) =.?
v² = u² – 2gh (since the ball is going against gravity)
0² = u² – (2 × 9.8 × 45)
0 = u² – 882
Collect like terms
0 + 882 = u²
882 = u²
Take the square root of both side
u = √882
u = 29.7 m/s
Therefore, the ball must be thrown with a speed of 29.7 m/s.
B. Determination of the time spent by the ball in the air.
We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:
Maximum height (h) = 45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) to reach the maximum height =?
h = ½gt²
45 = ½ × 9. 8 × t²
45 = 4.9 × t²
Divide both side by 4.9
t² = 45/4.9
Take the square root of both side
t = √(45/4.9)
t = 3.03 s
Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:
Time (t) to reach the maximum height = 3.03 s
Time (T) spent by the ball in the air =?
T = 2t
T = 2 × 3.03
T = 6.06 s
Therefore, the ball spent 6.06 s in the air.
Answers
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,
Answer:
4833J
Explanation:
Length=0.777
mass=2.67
# rods= 5
ω=573 rpm--> rad/s
I=kgm^2
K=1/2(number of rods)(I)(ω)=J
I know it's very late, but hope this helps anyone else trying to find the answer.
Answers
Answer:
200
Explanation:
The computation of the impedance of the circuit is shown below:
Provided that
RMS voltage = 120 v
Frequency = 60.0 Hz
RMS current = 0.600 A
Based on the above information, the formula to compute the impedance is
where,
And,
Now placing these above values to the formula
So, the impedance of the circuit is
= 200
Answers
Final answer:
The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.
Explanation:
To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.
First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.
Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.
So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.
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1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?
Answers
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
Final answer:
Positive, negative, and zero terms in the energy equation. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.
Explanation:
To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:
- Positive terms:
- ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
- ΔUs - the change in elastic potential energy is positive if there is any stretch or compression in the system.
- ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
- ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.
- ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.
To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in gravitational potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.
The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the metabolic power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.
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