Answers
Answer:
v = 23.66 m/s
Explanation:
recall that one of the equations of motion may be expressed:
v² = u² + 2as,
Where
v = final velocity (we are asked to find this)
u = initial velocity = 0 m/s since we are told that it starts from rest
a = acceleration = 0.56m/s²
s = distance traveled = given as 500m
Simply substitute the known values into the equation:
v² = u² + 2as
v² = 0 + 2(0.56)(500)
v² = 560
v = √560
v = 23.66 m/s
Related Questions
A 2.50-nF parallel-plate capacitor is charged to an initial potential difference ΔVi = 100 V and is then isolated. The dielectric material between the plates is paper, with a dielectric constant of 3.7.
Requried:
a. How much work is required to withdraw the mica sheet?
b. What is the potential difference across the capacitor after the mica is withdrawn?
Answers
Explanation:
Formula to calculate the energy stored in a capacitor when it is filled with air is,
[tex] U_{1}=\frac{1}{2} C V_{1}^{2} [/tex]]
Here, [tex]U_{1}[/tex] is the energy stored in a capacitor when it is filled with air.
[tex]C[/tex] is the parallel plate capacitor.
[/tex]V_{\mathrm{i}}[/tex] is the initial potential difference.
Substitute [tex]2.00 \mathrm{nF}[/tex] for [tex]C[/tex] and [tex]100 \mathrm{V}[/tex] for [tex]V_{\mathrm{i}}[/tex] to find the [tex]U_{1}[/tex]
[tex] \begin{array}{c} U_{1}=\frac{1}{2}\left(2.00 \mathrm{nF}\left(\frac{10^{9} \mathrm{F}}{1 \mathrm{n} \mathrm{F}}\right)\right)(100 \mathrm{V})^{2} \\ =10^{-5} \mathrm{J} \end{array} [/tex]]
Formula to calculate the energy stored in a capacitor when it is filled with dielectric is,
A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matter it passes through. One such particle, initially traveling at 2.40 ✕ 106 m/s in a straight line, decreases in speed to 1.56 ✕ 106 m/s over a distance of 1.22 km.
(a) What is the magnitude of the force experienced by the muon?
(b) How does this force compare to the weight of the muon?
|F|/Fg =______
Answers
Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evporate
Answers
Answer:
Gain heat and evaporate
Explanation:
As water molecules are exposed to sunlight, they begin to heat up. This means that the molecules begin to jiggle faster, and as such take up less space. Since they take up less space they are less dense, and therefore more bouyant. This means that they begin to rise into the air, and evaporate. Hope this helps!
the mass of a brick is 4 kg, find the mass of water displaced by it when completely immersed in water.
Answers
Answer:
The correct answer is = 1.6
Explanation:
Density of water = 1000kg/m³ = d₁
Mass of brick = 4kg = m
Density of brick = 2.5 g/cm³ = 2.5 × 1000 =2500 kg/m³ = d₂
Volume of brick = m/d₂ = 4/2500 =16/10000 = 0.0016 L = v
Buoyant Force = v × d₁ × g (g= acceleration due to gravity =9.8m/s²)
= 0.0016 × 1000 × 9.8 = 15.68 Newtons
By the Archimedes' Principle, the buoyant force is equal to the weight of the liquid displaced by an object.
Weight of the water displaced=Buoyant Force
=Mass of water displaced × g,
as weight = mass × acceleration due to gravity
15.68= mass of brick × 9.8
15.68/9.8 =Mass of water displaced
1.6 kg = Mass of water displaced
In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
ОА.
in the direction of the applied force
OB.
in the direction opposite of the direction of the applied force
OC. perpendicular to the direction of the applied force
OD
in a circular motion
Answers
Answer:
in the direction of the applied force
Explanation:
Coherent light that contains two wavelengths, 660 nmnm (red) and 470 nmnm (blue), passes through two narrow slits that are separated by 0.490 mmmm. Their interference pattern is observed on a screen 4.50 mm from the slits.Required:What is the distance on the screen between the first order bright fringes for the two wavelengths?
Answers
Answer:
[tex]1.94\times10^{-3}[/tex] m
Explanation:
Condition for constructive interference is
[tex]y =\frac{m\lambda}{d} D[/tex]
y= width of the first bright fringe
λ= wavelength of the incident light
d= distance between the slits
D= distance of the screen from the slit
for first order 1st wavelength
[tex]y_1 =\frac{1\times660\times10^{-9}}{0.49\times10^{-3}} 5[/tex]
[tex]y_1=6.73\times10^{-3} m[/tex]
Now, for first order 2nd wavelength
[tex]y_2 =\frac{1\times470\times10^{-9}}{0.49\times10^{-3}} 5[/tex]
[tex]y_2=4.79\times10^{-3} m[/tex]
The distance between the first bright fringe for each wavelength
[tex]d=y_1-y_2\\=(6.73-4.79)\times10^{-3} m\\=1.94\times10^{-3} m[/tex]
Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?
Answers
Answer:
a) E = 2.7x10⁶ N/C
b) F = 54 N
Explanation:
a) The electric field can be calculated as follows:
[tex] E = \frac{Kq}{d^{2}} [/tex]
Where:
K: is the Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
d: is the distance
Now, we need to find the electric field due to charge 1:
[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]
The electric field due to charge 2 is:
[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]
The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):
[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]
Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.
b) The force on a charge q₃ situated there is given by:
[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]
[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]
Therefore, the force on a charge q₃ situated there is 54 N.
I hope it helps you!
(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].
(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.
The answer can be explained as follows.
Electric FieldGiven that the two charges are;
[tex]q_1 = 50\times 10^{-6}\,C[/tex] and [tex]q_2 = -25\times 10^{-6}\,C[/tex](a) At the midpoint; [tex]r = 0.5\,m[/tex].
We know that the electric field due to charge [tex]q_1[/tex].
[tex]E_1 = k\,\frac{q_1}{r^2}[/tex]Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]
[tex]E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C[/tex]The electric field due to charge [tex]q_2[/tex] is given by;
[tex]E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C[/tex]Therefore, the net electric field in the midpoint is given by;
[tex]E_{net} =E_2+E_1[/tex][tex]\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C[/tex]The direction is towards the right side.
Electrostatic Force(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.
So the force on the charge is ;
[tex]F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N[/tex]Find out more about electrostatic force and fields here:
https://brainly.com/question/14621988
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.7 cm farther apart, the voltage between the plates increases by 100 V.(a) What is the charge Q on the positive plate of the capacitor?_________nC(b) How much does the energy stored in the capacitor increase due to the movement of the plates?_________µJ
Answers
Answer:
a) Q = 0.759µCb) E = 39.5µJExplanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
A) The charge Q on the positive plate of the capacitor is ; 0.759 µC
B) The energy stored in the capacitor increases by : 39.5 µJ
Given data :
Area of plates ( A ) = 600 cm²
Distance between plates ( d ) = 0.7 cm
Voltage across plates = 100 v
∈ ( permittivity of free space ) = 8.85 * 10⁻¹²
A) Determine the Charge on the positive plate of the capacitor
Q = CV --- ( 1 )
where ; C = ∈ * A / d and V = 100 v
∴ C = 8.85 * 10⁻¹² * 600 / 0.7 = 7.59 *10⁻⁹ F
Back to equation ( 1 )
Q = 7.59 *10⁻⁹ * 100
= 0.759 µC
B) Calculate how much The energy stored in the capacitor increases
E = 1/2 * C * V²
= 1/2 * 7.59 *10⁻⁹ * 100²
= 39.5 µJ
Hence we can conclude that The charge Q on the positive plate of the capacitor is ; 0.759 µC, The energy stored in the capacitor increases by : 39.5 µJ.
Learn more : https://brainly.com/question/1457596
4. Chloe has a vertical velocity of 3 m/s when she leaves the 1 m diving board. At this instant, her center of gravity is 2.5 m above the water. How high above the water will Chloe go
Answers
Answer:
2.95m
Explanation:
Using h= 2.5+ v²/2g
Where v= 3m/s
g= 9.8m/s²
h= 2.95m
Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish really is? Does the fish see the person's face closer or farther than it really is? Explain your answer.
Answers
Answer:
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Explanation:
This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.
In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer
1 sin θ₁ = 1.33 sin θ₂
θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)
In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away
Answer:
The fish appears closer than it really is because light from the fish is refracted away from the normal as it enters the air pocket in the goggles. This is because air has a smaller index of refraction than water. The person will trace rays back to an image point in front of the actual fish. The fish will see the person's face exactly where it actually is because the light from the face is not refracted as it travels through water only, and does not change from one medium to another.
Explanation:
Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t = 0 and subsequently experiences a constant angular acceleration α = 1.3 rad/s2. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 1.7 s.
Answers
Given that,
Angular velocity = 0.17 rad/s
Angular acceleration = 1.3 rad/s²
Time = 1.7 s
We need to calculate the angular velocity
Using angular equation of motion
[tex]\omega=\omega_{0}+\alpha t[/tex]
Put the value in the equation
[tex]\omega=0.17+1.3\times1.7[/tex]
[tex]\omega=2.38(k)\ m/s[/tex]
We need to calculate the angular displacement
Using angular equation of motion
[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]
Put the value in the equation
[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]
[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]
[tex]\theta= 124.18^{\circ}[/tex]
We need to calculate the velocity at point A
Using equation of motion
[tex]v_{A}=v_{0}+\omega\times r[/tex]
Put the value into the formula
[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]
[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]
[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]
We need to calculate the acceleration at point A
Using equation of motion
[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]
Put the value in the equation
[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]
[tex]a_{A}=0.791j-0.851i[/tex]
[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]
Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]
(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]
A proton moves at a speed 1.4 × 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.85 m. What is the field strength?
Answers
0.17T
Explanation:
When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e
[tex]F_{C}[/tex] = [tex]F_{M}[/tex] --------------(i)
But;
[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex] [m = mass of the particle, r = radius of the path, v = velocity of the charge]
[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]
Substitute these into equation (i) as follows;
[tex]\frac{mv^2}{r}[/tex] = qvB
Make B subject of the formula;
B = [tex]\frac{mV}{qr}[/tex] ---------------(ii)
Known constants
m = 1.67 x 10⁻²⁷kg
q = 1.6 x 10⁻¹⁹C
From the question;
v = 1.4 x 10⁷m/s
r = 0.85m
Substitute these values into equation(ii) as follows;
B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]
B = 0.17T
Therefore, the magnetic field strength is 0.17T
A 2kg block is sitting on a hinged ramp such that you can increase the angle of the incline. The coefficient of static friction between the block and the ramp is 0.67 and the coefficient of kinetic friction is 0.25.
a. What angle do you have to tilt the ramp to get the block to slide?
b. What acceleration does the block experience at this angle when kinetic friction takes over?
Answers
Answer:
θ = 33.8
a = 3.42 m/s²
Explanation:
given data
mass m = 2 kg
coefficient of static friction μs = 0.67
coefficient of kinetic friction μk = 0.25
solution
when block start slide
N = mg cosθ .............1
fs = mg sinθ ...............2
now we divide equation 2 by equation 1 we get
[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]
[tex]\frac{\mu s N }{N}[/tex] = tanθ
put here value we get
tan θ = 0.67
θ = 33.8
and
when block will slide then we apply newton 2nd law
mg sinθ - fk = ma ...............3
here fk = μk N = μk mg cosθ
so from equation 3 we get
mg sinθ - μk mg cosθ = ma
so a will be
a = (sinθ - μk cosθ)g
put here value and we get
a = (sin33.8 - 0.25 cos33.8) 9.8
a = 3.42 m/s²
An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answers
Answer:
2P
Explanation:
See attached file
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber.
1. In which case is the change in momentum of the ball between the instant just before the ball collides with the floor or rubber and the instant just after the ball leaves the floor or rubber the biggest?
a. Case 1
b. Case 2
c. Same in both
2. In which case is the average force acting on the ball during the collision the biggest?
a. Case 1
b. Case 2
c. Same in both
Answers
Answer:
1. c. Same in both
2. a. Case 1
Explanation:
1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.
2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.
What must the charge (sign and magnitude) of a 1.60 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 680 N/C
Answers
Answer:
Explanation:
The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .
Let the charge be - q .
force on charge
= q x E where E is electric field
= q x 680
weight = 1.6 x 10⁻³ x 9.8
so
q x 680 = 1.6 x 10⁻³ x 9.8
q = 1.6 x 10⁻³ x 9.8 / 680
= 23 x 10⁻⁶ C
- 23 μ C .
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube
Answers
Answer:
f = 357.29Hz
Explanation:
In order to calculate the fundamental frequency in the closed tube, you use the following formula:
[tex]f_n=\frac{nv}{4L}[/tex] (1)
n: order of the mode = 1
v: speed of sound = 343m/s
L: length of the tube = 24cm = 0.24m
You replace the values of the parameters in the equation (1):
[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]
The fundamental frequency of in the tube is 357.29Hz
I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answers
Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
Which of the following technologies is based on the work of Ibn al-Haytham?
A. Telescopes to observe the visible light of distant stars
B. Radiation treatments for breast cancer
C. Radar to detect the movement of storms
O D. An orbiting observatory to detect X-rays from space objects
Answers
Answer:
The answer is A
Explanation:
Its A because he created a telescope to be able to observe stars.
in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
plz help me out
I will mark u as brainliest if u answer correct
Answers
Answer:
The correct answer is option 3 .
Please check the answer once :)
f the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in the string at the top of the circle
Answers
Answer:
the size are components relative to the whole.
Explanation:
they are particularly good at showing percentage or proportional data
Objects floating in the water, like buoys, only bob up and down when waves pass. Why do they not get pushed all the way to wherever the wave goes
Answers
Answer:
Because as the waves propagates, the particles of the medium (molecules of water) vibrates perpendicularly (upward and downward) about their mean position and not in the direction of the waves.
Explanation:
A wave is a phenomena which causes a disturbance in a medium without any permanent deformation to the medium. Examples are; transverse wave and longitudinal wave. Waves transfer energy from one point in the medium to another.
The waves generated by water are transverse waves. Which are waves in which the vibrations of the particles of the medium is perpendicular to the direction of propagation of the waves.
Thus as the waves propagates, the molecules of water vibrates up and down and not along the direction of propagation of the waves. So that the floating objects do not get pushed in the direction of the waves every time.
Value of g in CGS system
Answers
Answer:
in CGS system G is denoted as gram
A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the electric field between the plates
Answers
Answer:
Majorly the electric field is reduced among other effect listed in the explanation
Explanation:
In capacitors the presence of di-electric materials
1. decreases the electric fields
2. increases the capacitance of the capacitors.
3. decreases the voltage hence limiting the flow of electric current.
The di-electric material serves as an insulator between the metal plates of the capacitors
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
Answers
Answer:
D. All of the above
Explanation:
The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.
James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure, moratorium, and achievement.
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 971 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed.
You find that the forces are attractive and the magnitude of each force is:______
Answers
Answer:
The magnitude of each force is 2.45 x 10⁻¹⁶ N
Explanation:
The charge of proton, +q = 1.603 x 10⁻¹⁹ C
The charge of electron, -q = 1.603 x 10⁻¹⁹ C
Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m
Apply Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
q₁ and q₂ are the charges of proton and electron respectively
F is the magnitude of force between them
Substitute in the given values and solve for F
[tex]F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N[/tex]
Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N
A ball is thrown at an angle 40.00 above the horizontal with an initial velocity of 22.0 m/s. What is the range of the ball?
Answers
Answer:
48.64 m
Explanation:
From the question,
Range(R) = (U²Sin2Ф)/g.................. Equation 1
Where U = initial velocity, Ф = Angle to the horizontal, g = acceleration due to gravity.
Given: U = 22 m/s, Ф = 40°, g = 9.8 m/s².
Substitute these values into equation 1
R = 22²Sin(40×2)/9.8
R = 484×0.9848/9.8
R = 48.64 m
Hence the range of the ball is 48.64 m
The electric field at point P due to a point charge Q a distance R away has magnitude E. In order to double the magnitude of the field at P, you could:__________.
a. double the distance to 2.
b. double the charge to 2 and at the same time reduce the distance to /2.
c. reduce the distance to /2.
d. reduce the distance to /4.
e. double the charge to 2.
Answers
Answer:
Option E, double the charge to 2, is the right answer.
Explanation:
Given the electric field at point = P
The point charge = Q
Distance =R
Magnitude = E
Due to a certain charge, the magnitude of the electric field at a point is defined as:
[tex]E = \frac{kQ}{r^2}[/tex]
Here, we can see that E is proportionate to Q and 1/r^2
Hence, double the charge to 2 and at the same time reduce the distance to /2 will half the E
Therefore, doubling the charge will doubling E
So the answer is (e) is correct.
Run 2 17. Set # of slits to 2 18. Set Wave Length to 400nm 19. Set Slit width to 1600 nm 20. Set Slit spacing to 5000nm In row 18 21. Record distance to 1st bright fringe 22. Record distance to 2nd bright fringe 23. Record distance to 3rd bright fringe Knowing the screen distance to be 1m 24. Calculated the measured angle to 1st bright fringe 25. Calculated the measured angle to 2nd bright fringe 26. Calculated the measured angle to 3rd bright fringe Using sin(θ)=mλ/d 27. Calculate θ for 1st bright fringe
Answers
Answer:
a) m=1, y₁ = 0.08 m , θ₁ = 4.57º , b) m=2, y₂ = 0.16 m , θ₂ = 9.09º , c) m=3, y₃ = 0.24 m , θ₃ = 13.5º
Explanation:
After reading your strange statement, I understand that this is an interference problem, I transcribe the data to have it more clearly. Number of slits 2, distance between slits 5000 nm, wavelength 400 nm, distance to the screen 1 m.
They ask us to calculate the angles for the first, second and third interference, they also ask us to write down the distance from the central maximum.
The expression for constructive interference for two slits is
d sin θ = m λ
where d is the distance between the slits, λ is the wavelength used, m is an integer representing the order of interference
Let's use trigonometry to find the distance from the central maximum
tan θ = y / L
in all interference experiments the angle is small,
tan θ = sin θ / cos θ = sin θ
sint θ = y / L
let's replace
d y / L = m λ
y = m λ L / d
let's calculate
distance to the first maximum m = 1
y₁ = 1 400 10⁻⁹ 1/5000 10⁻⁹
y₁ = 0.08 m
distance to second maximum m = 2
y₂ = 2 400 10⁻⁹ 1/5000 10⁻⁹
y₂ = 0.16 m
distance to the third maximum m = 3
y₃ = 3 400 10⁻⁹ 1/5000 10⁻⁹
y₃ = 0.24 m
with these values we can search for each angle
tan θ = y / L
θ = tan⁻¹ y / L
for m = 1
θ₁ = tan⁻¹ (0.08 / 1)
θ₁ = 4.57º
for m = 2
θ₂ = tan⁻¹ (0.16 / 1)
θ₂ = 9.09º
for m = 3
θ₃ = tan⁻¹ (0.24 / 1)
θ₃ = 13.5º
In Einstein's Thoery of Relativity. What did he believe was the relationship between energy and malter?
Answers
Explanation:
Einsteins theory of relativity explains how space and time are linked for objects that are moving at a consistent speed in a straight line.