Answer:
5. mv, 2mv
Explanation:
for clay :
initial momentum = P_i = - mv
final momentum = P_f = 0
change in momentum = Pf - Pi = 0 - (- mv) = mv
for ball :
initial momentum = P_i = - mv
final momentum = P_f = mv
change in momentum = Pf - Pi = mv - (-mv) = 2mv
so mv and 2mv are the change in momentum of the clay and ball, respectively
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
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
Where are metals generally found on the periodic table ?
Answer:
on the left of the periodic table
Answer:
in the middle towards the left
Explanation:
just finished the quiz
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
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.
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?
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
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?
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
Value of g in CGS system
Answer:
in CGS system G is denoted as gram
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
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.
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. What is the magnitude of the velocity of the shell as it hits the ground?
Answer:
[tex]V = 331.6946\ m/s[/tex]
Explanation:
First let's find the time that takes the shell to hit the ground (height zero).
To find this time, we can use the equation:
[tex]S = So + Vo*t + at^2/2[/tex]
Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time. Then, for the vertical movement of the shell, we have that:
[tex]0 = 80 + 0*t - 9.81*t^2/2[/tex]
[tex]9.81*t^2/2 = 80[/tex]
[tex]t^2 = 160/9.81 = 16.31[/tex]
[tex]t =4.0386\ seconds[/tex]
Now, to find the horizontal speed, we use the equation:
[tex]S = So + V*t[/tex]
Then, for the horizontal movement, we have:
[tex]1330 = 0 + V_h * 4.0386[/tex]
[tex]V_h = 1330/4.0386 =329.32\ m/s[/tex]
Now we need to find the vertical speed, using:
[tex]V = Vo + a*t[/tex]
[tex]V_v = 0 - 9.81*4.0386[/tex]
[tex]V_v = -39.6187\ m/s[/tex]
Finally, to find the magnitude of the velocity, we have:
[tex]V = \sqrt{V_h^2 + V_v^2}[/tex]
[tex]V = \sqrt{329.32^2 + (-39.6187)^2}[/tex]
[tex]V = 331.6946\ m/s[/tex]
the mass of a brick is 4 kg, find the mass of water displaced by it when completely immersed in water.
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
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?
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,
Match each term with the best description.
1. Ratio of the speed of light in a vacuum to the speed of light in a medium.
2. Rule for how light is refracted at the boundary between two materials.
3. Process that occurs when the angle of incidence is greater than the critical angle.
4. Path of a particle of light.
5. The separation of light into different frequencies to produce a rainbow.
A. Dispersion.
B. Ray.
C. Refractive index.
D. Snell's law.
E. Total internal reflection.
Answer:
Explanation:
1.) Ratio of the speed of light in a vacuum to the speed of light in a medium is called refractive index. It is also known as index of refraction. It is a measure of light ray as they travel from one medium to another medium. We can determine the nature of a pair of media through their refractive indices.
2) Rule for how light is refracted at the boundary between two materials is known as the Snell's law. Snell's law states that the ratio of the angle of incidence to the angle of refraction is a constant for a given pair of media. The angle incidence is greater than the angle of refraction if the light ray travels from less dense medium to a denser medium and lesser than the angle of refraction if the light ray travels from denser medium to a less dense medium.
3) Process that occurs when the angle of incidence is greater than the critical angle is known as Total internal reflection. Total internal reflection only occurs when light rays travels from a denser medium to a less dense medium.
4.) Path of a particle of light is known as Ray. A ray is a particle of light that emerge from the sun or a reflecting (luminous) body. They are in form of straight lines striking the surface of a body.
5.) Dispersion is defined as the separation of white light into into component color. This is accompanied with the use of a triangular glass prism. The white light is an example of electromagnetic spectrum. The component colors are known as ROYGBIV (Red, Orange, Yellow, Green, Blue, Indigo and violet.). This combination of colors are called rainbow.
The separation of light into different frequencies to produce a rainbow is therefore known as DISPERSION
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
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 .
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.
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:
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
Answer:
the size are components relative to the whole.
Explanation:
they are particularly good at showing percentage or proportional data
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:______
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
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?
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]
An airplane propeller is 2.18 m in length (from tip to tip) with mass 97.0 kg and is rotating at 2600 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod.What is its rotational kinetic energy?
Answer:
1.4*10^6 J
Explanation:
Given that
Length of the propeller, l = 2.18 m
Mass of the propeller, m = 97 kg
Speed of the propeller, w = 2600 rpm
The formula for finding rotational Kinetic energy, K is = ½Iw²
Where, I is the moment of Inertia, and is given as 1/12 * m * l²
I = 1/12 * 97 * 2.18²
I = 8.083 * 4.7524
I = 38.41 kgm²
w = 2600 rpm, converting to rad/s, we have
w = 2600 * 2π rad/s
w = 272.31 rad/s
Now, Kinetic Energy, K is
K = ½Iw²
K = ½ * 38.41 * 272.31²
K = 19.205 * 74152.7361
K = 1424103.3 J
K = 1.4 MJ or 1.4*10^6 J
Thus, the rotational Kinetic Energy is 1.4*10^6 J
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.
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.
In Einstein's Thoery of Relativity. What did he believe was the relationship between energy and malter?
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.
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?
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
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
Answer:
The answer is A
Explanation:
Its A because he created a telescope to be able to observe stars.
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
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
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
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
the mass of a brick is 4 kg, find the mass of water displaced by it when completely immersed in water.
Answer:
Mass of water displaced = 1.6 kg
Explanation:
given data
Mass of brick m = 4kg
Density of water d₁ = 1000kg/m³
Density of brick d₂ = 2.5 g/cm³ = 2500 kg/m³
solution
we get here first volume of brick that is
volume of brick = m ÷ d₂ ..................1
put here value and we get
volume of brick = 4 ÷ 2500
volume of brick = 0.0016 L
and
we will use here Archimedes' Principle formula of the buoyant force
as buoyant force = the weight of the liquid displaced by an object
so now we get here Buoyant Force that is express as
Buoyant Force = v × d₁ × g ........................2
put here value
Buoyant Force = 0.0016 × 1000 × 9.8
Buoyant Force = 15.68 Newtons
so
Buoyant Force = Weight of the water displaced .................3
Buoyant Force = Mass of water displaced × g
and
weight = mass × acceleration due to gravity .......................4
so put here value and we will get
15.68 = mass of brick × 9.8
Mass of water displaced = 1.6 kg
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
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.
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.
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]
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
Answer:
in the direction of the applied force
Explanation:
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
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.
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
Answer:
The correct answer is option 3 .
Please check the answer once :)
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
Answer:
2.95m
Explanation:
Using h= 2.5+ v²/2g
Where v= 3m/s
g= 9.8m/s²
h= 2.95m