Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, [tex]$\omega = \frac{1}{8}$[/tex] rev/sec
[tex]$=\frac{2 \pi \times 7.5}{8}$[/tex] rad/sec
= 5.89 rad/sec
Therefore, force required,
[tex]$F=m.\omega^2.r$[/tex]
[tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
water contracts on freezing is it incorrect or conrrect
Answer:
hope it helps
much as you can
why do you like the full moon ?
Answer:
The Moon brings perspective. Observing the Moon, and I mean really looking – sitting comfortably, or lying down on a patch of grass and letting her light fill your eyes, it's easy to be reminded of how ancient and everlasting the celestial bodies are. When I do this, it always puts my life into perspective.Answer:
because it look more impressive than empty dark sky .
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Define measurements.
Answer:
act or process of measuring
Explanation:
Explanation:
the comparison of an unknown quantity with a known quantity.
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
A small plane tows a glider at constant speed and altitude. If the plane does 2.00 * 105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the tow rope and the horizontal
Answer:
θ = 57.4°
Explanation:
The complete formula to find out the work done by the plane is as follows:
[tex]W = FdCos\theta[/tex]
where,
W = Work = 200000 J
F = Force = Tension = 2560 N
d = distance = 145 m
θ = angle between rope and horizontal = ?
Therefore,
[tex]200000\ J = (2560\ N)(145\ m)Cos\theta\\\\Cos\theta = \frac{200000\ J}{371200\ J}\\\\\theta = Cos^{-1}(0.539)[/tex]
θ = 57.4°
The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Complete question:
A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.
Answer:
The tension on the rope is 1.95 N
Explanation:
The general equation of a progressive wave is given as;
[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]
Compare the given equation with the general equation of wave, the following parameters will be deduced.
A = 0.75 cm
k = 0.400π cm⁻¹
ω = 250π s⁻¹
The frequency of the wave is calculated as;
ω = 2πf
2πf = 250π
2f = 250
f = 250/2
f = 125 Hz
The wavelength of the wave is calculated as;
[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]
The velocity of the wave is calculated as;
v = fλ
v = 125 x 0.05
v = 6.25 m/s
The tension on the rope is calculated as;
[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]
Therefore, the tension on the rope is 1.95 N
Help me plssssssss cause I’m struggling
Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
why is unit of power is called derived unit?
Distance travelled by a body in unit time is called speed. it is a scalar quantity because it can be specified only by magnitude.
a. Give an example of the conversion of light energy to electrical energy.
b. Give an example of chemical energy converting to heat energy.
c. Give an example of mechanical energy converting to heat energy.
Explanation:
a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy
b) burning of coal converts chemical energy to heat energy
c) rubbing of both hands against each other converts mechanical to heat energy
Answer:
a. solar cells
b.coal,wood,petroleum
c.rubbing ours palms
Name the electrolyte in the chemical method of generating electricity
A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips
Answer:
I could not find the answer or do it myself if I did find it I would defenetly share
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0 m and Ay 450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0, but the magnitude (450.0 m) and direction of vector remain unchanged, as in drawing b. What are the scalar components, Ax and Ay, of the vector in the rotated x, y coordinate system
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
how did kepler discoveries contribute to astronomy
Answer:
They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.
Explanation:
The current in a conductor is 2.5A .explain the meaning of this statement
Answer:
In 1 second the amount of charge flowing through the conductor is 2.5 Q.
Explanation:
Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.
Answer:
Explanation:
English alphabets numbered fro 1 to 26
and the numbers 1 to10 so they are written in roman numbers as
1 - I
2 - II
3 - III
4 - IV
5 -V
6 - VI
7 -VII
8 - VIII
9 - IX
10 -X
11 - XI
12 - XII
13 - XIII
14 - XIV
15 - XV
16 - XVI
17 - XVII
18 - XVIII
19 - XIX
20- XX
21 - XXI
22 - XXII
23 - XXIII
24 - XXIV
25 - XXV
26 - XXVI
Which level of government relies the most on income tax?
OA.
federal
state
OC.
local
Answer:
Its the Federal government
lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.360 s, during which it produces an average 0.690 W from an average 3.00 V. (a) What energy does it dissipate
Energy = (power) x (time)
Energy = (0.69 W) x (0.36 sec)
Energy = 0.25 Joule
The pressure exerted at the bottom of a column of liquid is 30 kPa. The height of the
column is 3,875 m. What type of liquid is used?
Answer:
For example, the pressure acting on a dam at the bottom of a reservoir is ... pressure = height of column × density of the liquid × gravitational field ... The density of water is 1,000 kg/m 3.
1.Lõi thép máy biến áp được ghép từcác lá thép là để:
(a) Giảm tổn hao công suất do dòng điện xoáy
(b) Giảm tổn hao công suất do từ trễ
(c) Giảm tổn hao công suất do dòng điện chạy qua dây quấn
(d) Giảm tất cảcác loại tổn hao công suất.
Answer:
Option (c)
Explanation:
1.The transformer core is assembled from steel sheets to:
(a) Reduced power loss due to eddy current
(b) Reduced power loss due to hysteresis
(c) Reduced power loss due to current flowing through the winding
(d) Reduce all types of power loss.
A transformer is a device which converts the low voltage into high and vice versa.
There are two types of a transformer.
Step up: It is used to convert low voltage into high.
Step down It is used to convert high voltage into high.
It depends on the number of turns in primary and the secondary coil.
The core of the transformer is laminated and it is in the form of sheets.
By using such type of core, the power loss due to the windings is reduced.
option (c) .
~~~~~NEED HELP ASAP~~~~~
A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)
a.) Determine the linear regression
b.) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
a concrete has a height of 5m and has unit area 3m² supports a mass of 30000kg.
Determine the stress, strain and change in height
Answer:
stress = 98000 N/m^2
strain = 3.92 x 10^-6
change in height = 0.0196 mm
Explanation:
Height, h = 5 m
Area, A = 3 m²
mass, m = 30000 kg
Stress is defined as the force per unit area.
[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]
Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2
Young's modulus is defined as the ratio of stress to the strain.
[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]
let the change in height is h'.
Strain is defined as the ratio of change in height to the original height.
[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]
~~~~NEED HELP ASAP~~~~
Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.
Block A= 3kg
Block B= 4kg
Block A velocity before collision =30m/s.
Block B velocity before collision = 15 m/s
The velocity of block B after the collision is 20m/s.
a.) What is the velocity of block A after collision?
b.) Is the collision elastic? Show work to explain answer why or why not.
Answer:
Block A velocity is 23.33 m/s and the collission is not elastic.
Explanation:
a) m1v1 + m2v2 = m1v1' + m2v2'
Plug in givens
90+60=3v1'+80
solve for v1'= 23.33m/s
b) Find the initial and final kinetic energy of Block B
Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J
Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J
Since Ki does not equal Kf the collision is not elastic
The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Explanation:
Let T is the period of a pendulum. The SI unit of time is seconds (s).
It depends on the acceleration of gravity, g, and the length of the pendulum, l.
The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.
If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.
So,
[tex]T\propto \sqrt{\dfrac{l}{g}}[/tex]
Hence, this is the required solution.
Two substances, M and N, have specific heats c and 2c. if heats Q and 4Q are supɔlied to Mand N, respectively, their changes in temperature become equal. If substance M has mass m, find the mass of substance N in terms of m
Answer:
If the mass of B is m and the temperature change is the same, the mass of B will be 2m.
Explanation:
Q = mcT
T = mc/Q
M = 4Q/2cT........... (1)
T = Q/mc
Plug this in equation 1.
M = 4Q/(2c × Q/mc) = 4Q ÷ 2Q/m = 4Q × m/2Q = 2m
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Answer:
[tex]\lambda=6.83\times 10^{-5}\ m[/tex]
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]
So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].
A 10 kg box is at static equilibrium and the downward pull of gravity acting on the box is 98 Newton’s what is the minimum force that would require to just pick up the box
Explanation:
static equilibrium means its on the floor or something
so slightly greater than 98 newtons in the upward direction
Which phase of matter makes up stars?
O liquid
O gas
O plasma
Answer:
The answer to this question is plasma
Answer:
Plasma
Explanation:
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car B is 100 from car A, car A begins to accelerate toward car B with a constant acceleration of 5 m/s/s. Let right be positive.
1) How much time elapses before the two cars meet? 2) How far does car A travel before the two cars meet? 3) What is the velocity of car B when the two cars meet?
4) What is the velocity of car A when the two cars meet?
Answer:
Let's define t = 0s (the initial time) as the moment when Car A starts moving.
Let's find the movement equations of each car.
A:
We know that Car A accelerations with a constant acceleration of 5m/s^2
Then the acceleration equation is:
[tex]A_a(t) = 5m/s^2[/tex]
To get the velocity, we integrate over time:
[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]
Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:
[tex]V_a(t) = (5m/s^2)*t[/tex]
To get the position equation we integrate again over time:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]
Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:
[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]
Now let's find the equations for car B.
We know that Car B does not accelerate, then it has a constant velocity given by:
[tex]V_b(t) =20m/s[/tex]
To get the position equation, we can integrate:
[tex]P_b(t) = (20m/s)*t + P_0[/tex]
This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:
[tex]P_b(t) = (20m/s)*t + 100m[/tex]
Now we can answer this:
1) The two cars will meet when their position equations are equal, so we must have:
[tex]P_a(t) = P_b(t)[/tex]
We can solve this for t.
[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]
This is a quadratic equation, the solutions are given by the Bhaskara's formula:
[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]
We only care for the positive solution, which is:
[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]
Car A reaches Car B after 11.48 seconds.
2) How far does car A travel before the two cars meet?
Here we only need to evaluate the position equation for Car A in t = 11.48s:
[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]
3) What is the velocity of car B when the two cars meet?
Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s
4) What is the velocity of car A when the two cars meet?
Here we need to evaluate the velocity equation for Car A at t = 11.48s
[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]
