Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
Một vô lăng sau khi bắt đầu quay được một phút thì thu được vận tốc 700
vòng/phút. Tính gia tốc góc của vô lăng
1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
following:
blue chip
b. yellow chip
c. yellow chip
Answer:
Explanation:
Blue: 10/30
Red: 5/30
Yellow: 15/30
The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
What is the probability?
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
Given that
Blue chips =10
Red chips = 5
Yellow chips = 15
Number of the total samples =10+15+5=30
Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]
Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]
Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]
Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
To know more about probability follow
https://brainly.com/question/24756209
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 mm, and a baseball has a mass of 145 g.
Required:
a. Draw a free-body diagram of the ball during the pitch.
b. What force did the pitcher exert on the ball during this record-setting pitch?
c. Estimate the force in part b as a fraction of the pitcher's weight.
Answer:
Following are the solution to the given points:
Explanation:
For point a:
Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:
The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.
For point b:
[tex]160.2\ N[/tex]
First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that
where
[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]
Calculating the mass:
[tex]m = 145 g = 0.145 kg[/tex]
Calculating the force:
[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]
For point c:
0.195 times the pitcher's weight
[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]
Solving for W:
[tex]W=84 \times 9.8= 823.2 \ N[/tex]
Now the force of Part B could be defined as the fraction of the mass of the pitcher:
[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]
You should extending your throwing hand straight up to the sky to follow-through.
O True
O False
False
It's not straight up
2. Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute
Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)
A) 59 eV
B) 58 eV
C) 57 eV
D) 56 eV
Answer:
a. 59 ev. helpful answer
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Need an answer in hurry u can make the pic big
what does Newton's third law ? Describe.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.
Hope it helps
Answer:
Newton's third law state that every action there is equal but opposite reactions
hope this will help you more
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
which of the following is the correct description of momentum?
-the product of mass and acceleration -the product of mass and velocity
-velocity divided by mass
-acceleration divided by mass
Answer:
The product of mass and velocity is the correct answer
Explanation:
Momentum is defined as mass × velocity
p = mv
Answer:
The product of mass and velocity
Explanation:
I just did it and got it right with a 100%
Posted 1/3/23
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
djfjci3jsjdjdjdjdjddndn
ds
Need in hurry important please
Answer:
I don't see anything on your question?
A laser beam enters one of the sloping faces of the equilateral glass prism (n=1.42) and refracts through the prism. Within the prism the light travels horizontally. What is the angle between the direction of the incident ray and the direction of the outgoing ray?
Answer:
30.5°
Explanation:
Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from
n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).
Making D subject of the formula, we have
n = [sin(D + α)/2]/sin(α/2)
nsin(α/2) = [sin(D + α)/2]
(D + α)/2 = sin⁻¹[nsin(α/2)]
D + α = 2sin⁻¹[nsin(α/2)]
D = 2sin⁻¹[nsin(α/2)] - α
So, substituting the values of the variables into the equation, we have
D = 2sin⁻¹[nsin(α/2)] - α
D = 2sin⁻¹[1.42sin(60°/2)] - 60°
D = 2sin⁻¹[1.42sin(30°)] - 60°
D = 2sin⁻¹[1.42 × 0.5] - 60°
D = 2sin⁻¹[0.71] - 60°
D = 2(45.23°) - 60°
D = 90.46° - 60°
D = 30.46°
D ≅ 30.5°
A circular ice rink is 20 m in diameter and is to be temporarily enclosed in a hemispherical dome of the same diameter. The ice is maintained at 270 K. On a particular day the inner surface of the dome is measured to be 290 K. Estimate the radiant heat transfer from the dome to the rink if both surfaces can be taken as blackbody.
570k or 1050k
270k+290k= 570k
270k•290k = 1050k
A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how much work has she done?
Answer:
W = 354.9 J
Explanation:
Given that,
The mass of a bale of hay, m = 23 kg
The displacement, d = 3.9 m
The horizontal force exerted on the hay, F = 91 N
We need to find the work done. We know that,
We know that,
Work done, W = Fd
So,
W = 91 N × 3.9 m
W = 354.9 J
So, the required work done is 354.9 J.
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
P5. A bullet with an initial velocity of 280 m/s in the x-direction penetrates an initially stationary block of mass 11 kg and emerges on the other side with a final velocity of 70 m/s in the x-direction. The velocity of the block after the collision is 0.2 m/s, also in the x-direction. Assume the block slides on a horizontal frictionless surface. What is the mass of the bullet
Answer:
the mass of the bullet is 10.5 g
Explanation:
Given;
initial velocity, u₁ = 280 m/s
final velocity of the bullet, v₁ = 70 m/s
final velocity of the block, v₂ = 0.2 m/s
mass of the block, m₂ = 11 kg
initial velocity of the block, u₂ = 0
let the mass of the bullet = m₁
Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
280m₁ + 11(0) = 70m₁ + 11 x 0.2
280m₁ = 70m₁ + 2.2
280m₁ - 70m₁ = 2.2
210m₁ = 2.2
m₁ = 2.2/210
m₁ = 0.0105 kg
m₁ = 10.5 g
Therefore, the mass of the bullet is 10.5 g
I NEED THE ANSWER QUICK PLEASEE
in which states of matter will a substance have a fixed volume
Answer:
Solid is the state in which Matter maintains a fixed volume
Answer:
The state of matter that has a fixed volume is Solid.
Explanation:
Solid substances will maintain a fixed volume and shape.
A race car starts from rest on a circular track of radius 378 m. The car's speed increases at the constant rate of 0.580 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
a. The speed of the race car
b. The distance traveled
c. the elapsed time
Answer:
a) [tex]V=14.904m/s[/tex]
b) [tex]d = 191.49 m[/tex]
c) [tex]t= 25.696 s[/tex]
Explanation:
From the question we are told that:
Radius [tex]r =378m[/tex]
Acceleration [tex]a=0.580[/tex]
a)
Generally the equation for speed of the car is mathematically given by
[tex]a=\frac{v^2}{r}[/tex]
[tex]V=\sqrt{a*r}[/tex]
[tex]V=\sqrt{0.58*383}[/tex]
[tex]V=14.904m/s[/tex]
b)
Generally the equation for distance traveled of the car is mathematically given by
[tex]V^2=u^2+2ad[/tex]
[tex]d=\frac{V^2}{2a}[/tex]
[tex]d=\frac{14.904^2}{2*0.58}[/tex]
[tex]d = 191.49 m[/tex]
c)
Generally the equation for time of the car is mathematically given by
[tex]V=u+at[/tex]
[tex]t=\frac{V}{a}[/tex]
[tex]t=\frac{14.904}{0.58}[/tex]
[tex]t= 25.696 s[/tex]
2 coplas o pregones inventadas por ti relacionadas con la región caribe.
A toddler weighs 10 kg and raises herself onto tiptoe (on both feet). Her feet are 8 cm long with each ankle joint being located 4.5 cm from the point at which her feet contact the floor. While standing on tip toe:
(a) what is the upward normal force exerted by the floor at the point at which one of the toddler's feet contacts the floor?
(b) what is the tension force in one of her Achilles tendons? (c) what is the downward force exerted on one of the toddler's
ankle joints?
Answer:
a.49 n
b. 63 n
c. 112 n
Explanation:
a.10 times 9.8 from gravity/2 = 49 n
b. 49n times 4.5/8-4.5 = 63 n
c 49n + 63 n = 112 n
Consider different points along one spoke of a wheel rotating with constant angular velocity. Which of the following is true regarding the centripetal acceleration at a particular instant of time?
a. The magnitude of the centripetal acceleration is greater for points on the spoke closer to the hub than for points closer to the rim
b. both the magnitude and the direction of the centripetal acceleration depend on the location of the point on the spoke.
c. The magnitude of the centripetal acceleration is smaller for points on the spoke closer to the hub than for points closer to the rim but the direction of the acceleration is the same at all points on this spoke.
d. The magnitude and direction of the centripetal acceleration is the same at all points on this spoke.
Answer:
Option (a).
Explanation:
Let the angular velocity is w.
The centripetal acceleration is given by
[tex]a = r w^2[/tex]
where, r is the distance between the axle and the spoke.
So, more is the distance more is the centripetal acceleration.
(a) For the points on the spoke closer to the hub than for points closer to the rim is larger distance, so the centripetal force is more.
The statement is true.
(b) The direction of centripetal acceleration is always towards the center, so the statement is false.
(c) It is false.
(d) It is false.
Option (a) is correct.
Coherent light with wavelength 597 nm passes through two very narrow slits, and theinterference pattern is observed on a screen a distance of 3.00{\rm m} from the slits. The first-order bright fringe is adistance of 4.84 {\rm mm} from the center of the central bright fringe.
For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Answer:
The required wavelength is 1.19 μm
Explanation:
In the double-slit study, the formula below determines the position of light fringes [tex]y_m[/tex] on-screen.
[tex]y_m = \dfrac{m \lambda D}{d}[/tex]
where;
m = fringe order
d = slit separation
λ = wavelength
D = distance between screen to the source
For the first bright fringe, m = 1, and we make (d) the subject, we have:
[tex]d = \dfrac{(1) \lambda D}{y_1}[/tex]
[tex]d = \dfrac{ \lambda D}{y_1}[/tex]
replacing the value from the given question, we get:
[tex]d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m[/tex]
In the double-slit study, the formula which illustrates the position of dark fringes [tex]y_m[/tex] on-screen can be illustrated as:
[tex]y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
The value of m in the dark fringe first order = 0
∴
[tex]y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
[tex]y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}[/tex]
making λ the subject of the formula, we have:
[tex]\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}[/tex]
[tex]\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m[/tex]
Explain why energy cannot escape from the room by conduction
Answer: Heat energy transfer by conduction, convection and radiation
Heat energy is a very difficult energy to store as it can transfer in three different ways from warm surroundings to cooler surroundings. The three processes are conduction, convection or radiation.
Understanding energy, how it is transferred and how the amount of energy that is usefully transferred can be used to measure the efficiency is very important to physics and to the world.
how can the starch be removed from the leaves of potted plants
Answer:
Explanation:
There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.