Using second equation of kinematics
[tex]\boxed{\sf s=ut+\dfrac{1}{2}at^2}[/tex]
[tex]\\ \qquad\quad\sf{:}\dashrightarrow s=0(5)+\dfrac{1}{2}(8)(5)^2[/tex]
[tex]\\ \qquad\quad\sf{:}\dashrightarrow s=4(25)[/tex]
[tex]\\ \qquad\quad\sf{:}\dashrightarrow s=100m[/tex]
Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struck at the same time. A small piece of putty is placed on the tuning fork with the known frequency and it's frequency is lowered slightly. When struck at the same time, the two forks now produce a beat frequency of 8 Hz. 1)What is frequency of tuning fork which originally had a frequency of 335 Hz after the putty has been placed on it
Answer:
Explanation:
Unknown fork frequency is either
335 + 5.3 = 340.3 Hz
or
335 - 5.3 = 329.7 Hz
After we modify the known fork, the unknown fork frequency equation becomes either
(335 - x) + 8 = 340.3
(335 - x) = 332.3
x = 2.7 Hz
or
(335 - x) + 8 = 329.7
(335 - x) = 321.7
x = 13.3 Hz
IF the unknown fork frequency was 340.3 Hz,
THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz
IF the unknown fork frequency was 329.7 Hz,
THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz
A ball is thrown from the ground with velocity of 30m/s after what time has the ball has velocity of 10m/s down ward?
a ball is thrown vertically upward from the ground with the velocity of 30m/s. a) how long will it take to rise to its highest poitn? b) how high does the ball rise? c) how long after projection will the ball have a velocity of 10m/s? d) what is the total time of flight?
From Vf = Vo - gt, 0 = 30 - 9.8t yielding t = 3.06 sec.
From h = Vo(t) - g(t^2)/2, h = 30(3.06) - 4.9(3.06)^2 yielding h = 45.88m.
From Vn = Vo - gt, 20 - 9.8t yielding t = 2.04 sec.
Since the fall time equals the rise time, the total flight time is 2(3.06) = 6.12 sec.
Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After reflecting from the mirror, where will the rays cross each other at a single point?
The rays __________
a. will not cross each other after reflecting from a concave mirror.
b. will cross at the center of curvature.
c. will cross at the point where the principal axis intersects the mirror.
d. will cross at the focal point. will cross at a point beyond the center of curvature.
A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.
A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror. This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.
In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.
Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.
For reference: https://brainly.com/question/20380620
What particles in an atom can increase and decrease in number without changing the identity of the elements
Answer:
The number of neutrons or electrons in an atom can change without changing the identity of the element.
two object A and B vertically thrown upward with velocities of 80m/s and 100m/s at two seconds interview where and when will the two object meet.
Answer:
THIS IS YOUR ANSWER:
☺✍️HOPE IT HELPS YOU ✍️☺
Given the solution [tex]y_{1}(x)[/tex] from EDO below, develop a second solution.
[tex]x\frac{d^{2}y }{dx^{2} } +3\frac{dy}{dx} -y=0,\\y_{1} (x)=1+\frac{x}{3} +\frac{x^{2} }{24} +\frac{x^{3} }{360} + ...[/tex]
We're given
[tex]\displaystyle y_1(x) = 1 + \frac x3 + \frac{x^2}{24} + \frac{x^3}{360} + \cdots = \sum_{n=0}^\infty a_nx^n[/tex]
so let's see if we can find a closed form for the n-th term's coefficient.
Notice that
[tex]\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1\times3} \\\\ a_2 = \frac1{24} = \frac1{(1\times3) \times (2\times4)} \\\\ a_3 = \frac1{360} = \frac1{(1\times3) \times (2\times4) \times (3\times5)}[/tex]
If the pattern continues, the next few terms are likely
[tex]\displaystyle a_4 = \frac1{8640} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6)} \\\\ a_5 = \frac1{302400} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7) \times (6\times8)}[/tex]
which leads up to the n-th term,
[tex]\displaystyle a_n = \frac1{(1\times3) \times (2\times4) \times \cdots \times (n\times(n+2))} = \frac2{n!(n+2)!}[/tex]
where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.
So we have
[tex]\displaystyle y_1(x) = \sum_{n=0}^\infty \frac2{n!(n+2)!} x^n[/tex]
Now we use reduction of order to find a linearly independent solution of the form [tex]y_2(x) = v(x)y_1(x)[/tex], with derivatives
[tex]\displaystyle \frac{\mathrm dy_2}{\mathrm dx} = v(x) \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm dv}{\mathrm dx} \\\\ \frac{\mathrm d^2y_2}{\mathrm dx^2} = v(x) \frac{\mathrm d^2y_1}{\mathrm dx} + 2 \frac{\mathrm dv}{\mathrm dx} \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm d^2v}{\mathrm dx^2}[/tex]
Substitute [tex]y_2[/tex] and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :
[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} + \left(x\frac{\mathrm d^2y_1}{\mathrm dx^2}+3\frac{\mathrm dy_1}{\mathrm dx}-y_1\right)v = 0[/tex]
but since we know [tex]y_1(x)[/tex] satisfies the original DE, the last term vanishes and we're left with
[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} = 0[/tex]
Reduce the order by substituting [tex]w(x)=\dfrac{\mathrm dv}{\mathrm dx}[/tex] to get yet another DE in w(x) :
[tex]\displaystyle x y_1 \frac{\mathrm dw}{\mathrm dx} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)w = 0[/tex]
This equation is separable:
[tex]\displaystyle \frac{\mathrm dw}w = - \frac{2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1}{xy_1}\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}\frac{\mathrm dy_1}{\mathrm dx} + \frac3x\right)\,\mathrm dx[/tex]
From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for [tex]y_2(x)[/tex] by multiplying [tex]y_1(x)[/tex] by v(x). Using the fundamental theorem of calculus, you would find
[tex]\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = \frac{C_1}{x^3} \exp\left(-2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)} \,\mathrm d\xi\right)\right) \\\\ v = C_1 \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2[/tex]
so that you end up with
[tex]\displaystyle y_2(x) = C_1 y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2y_1(x)[/tex]
But the second term is already accounted for by [tex]y_1(x)[/tex] itself, so the second solution is
[tex]\displaystyle y_2(x) = \boxed{y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega}[/tex]
You could go the extra mile and try to find a power series expression for this solution, but that's a lot of work for little payoff IMO.
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
we will demonstrate experimentally that light travel in staight line.
OK.
Good luck on your experimental demonstration.
It's a nice exercise.
uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0 m from the base of the wall. What must be the magnitude of the force of static friction supplied by the floorto keep the ladder from slipping
Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
If a wave has speed of 235 m/s with a wavelength of 3 m, what is the frequency of the wave?
A racing car going a 20 m/s stops in a distance of 20 m.What is its acceleration?
step by step
formular v^2 = u^2+2as
stop v = 0
0 = 400+2a(20)
-400=40a
a = -10 m/s^2
ans affect acceleration is 10 m/s^2
No matter how far you stands from a mirror your image appear errect .the mirror is
Answer:
convex mirror
.....................
Answer:
convex mirror..........
Describe sound and record
Answer:
record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.
sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.
I hope this helps
what is liquid pressure and its si unit?
The SI unit of pressure is the pascal: 1Pa=1N/m2 1 Pa = 1 N/m 2 . Pressure due to the weight of a liquid of constant density is given by p=ρgh p = ρ g h , where p is the pressure, h is the depth of the liquid, ρ is the density of the liquid, and g is the acceleration due to gravity.
Find the acceleration of the blocks when the system is released. The coefficient of kinetic friction is 0.4, and the mass of each block is 1 kg. Neglect the mass of the pulleys and cord.
Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender to re-transmit the frame and hence it will increase the re-transmission overhead. By considering the scenarios given below, you have to choose whether the packets should be encapsulated in a single frame or multiple frames in order to minimize the re-transmission overhead.
Justify your answer with one valid reason for both the scenarios given below.
Scenario A: Suppose you are using a network which is very prone to errors.
Scenario B: Suppose you are using a network with high reliability and accuracy.
1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.
2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.
Justification:
There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors. The reliable and accurate network environment makes a single frame economically better.
Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.
With encapsulation, each layer:
provides a service to the layer above itcommunicates with a corresponding receiving nodeThus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead. This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.
Learn more about encapsulation of packets here: https://brainly.com/question/22471914
A circular loop of wire is in the plane of the paper. The south pole of a bar magnet is being moved from a position in front of the paper in a direction away from the center of the loop. The direction of the induced current in the loop. Which is the direction of the induced current in the loop?
Answer:
Counterclockwise
explanation in attachment
a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
Ibrah open a bottle of perfume infront of the room. After few minutes the smell of perfume reach the whole room. Explain why this happens
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s
Answer:
9 Brainly hahaha ............huh
Which columns are mislabeled?
Answer:
first order date and most recent order date
Explanation:
it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019
A 1.50 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s , and at point B it has slowed to 1.25 m/s .
Part A
How much work was done on the book between A and B ?
Part B
If -0.750J of work is done on the book from B to C , how fast is it moving at point C ?
Part C
How fast would it be moving at C if 0.750J of work were done on it from B to C ?
I assume friction is the only force acting on the book as it slides.
(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²
W ≈ -6.56 J
(B) Using the work-energy theorem again, the speed v of the book at point C is such that
-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v = 0.750 m/s
(C) Take the left side to be positive, then solve again for v.
0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²
==> v ≈ 1.60 m/s
what is angular frequency
Why can't cars be constructed that can magnetically levitate in Earth's magnetic field?
a. The current needed would be excessive.
b. The current within the car must flow in a complete circuit, about half Of which would exert forces against levitation.
c. Magnetic forces are always very weak.
d. Earth's magnetic field is in the wrong direction.
If a bus travels 50 km in 10 hours, how fast was the
bus travelling?
Answer:
5 kilometers per hour
Explanation:
Speed = distance / time
Distance: 50km
Time: 10 hours
Speed = 50/10 = 5kph
Answer:
5kmph
Explanation:
if the bus traveled 50 km in 10 hours, we have to divide 50 by 10 to see how fast it traveled per hour.
50/10 = 5
therefore, the bus was traveling 5 km per hour
hope this helps :)
g According to the Third Law, the action and reaction forces are exactly equal in magnitude and in opposite directions. So when a horse pulls a cart, the cart pulls the horse with the same force. Why then does the cart move
Answer:
One way to look at this is to consider the forces acting on any point in a string.
For a very small portion of string F = M a must still hold. As M approaches zero the small portion of string would have to approach infinite acceleration if the net force on that portion of string were not zero.
One generally considers the net force acting on the center of mass of an object not the individual forces acting on each infinitesimal mass composing
the object.
if an object weighs 550 n and the area is 1 cube
a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection
Answer:
Answer:30 cm
Answer:30 cmExplanation:
Answer:30 cmExplanation:Given=ROC= 60cm
Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.
An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside the pipe are
Answer:
The longest wavelength for closed at one end and open at the other is
y / 4 where y is the wavelength - that is node - antinode
The next possible wavelength is 3 y / 4 - node - antinode - node -antinode
y / 4 = 3 m y = 12 meters the longest wavelength
3 y / 4 = 3 m y = 4 meters 1 / 3 times as long
