Answer:
[tex]r_{cm}[/tex]= 1/5 L
Explanation:
To find the center of mass of the system let's use
[tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i
where m is the total mass of the system
let's apply this expression to our case
Let's set the reference frame on the star
[tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]
r_{cm} = [tex]\frac{1}{5}[/tex] L
[tex]r_{cm}[/tex]= 1/5 L
Hey, the breast center is 1/5 of the distance between the star and the planet.
Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10-m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope (5-m diameter) on Palomar Mountain in California
Answer:
Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Explanation:
If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.
Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
If acceleration is zero what statement about velocity is
true *
A)Velocity is zero
B)Velocity is constant
C)Velocity cannot be determined
D) Velocity is changing
Answer: A
Velocity is zero because the acceleration isn't affected, and velocity is the rate of change, so it can't be any other options.
Answer:
B)Velocity is constantExplanation:
If an object moves with a velocity and there is no acceleration, then the velocity remains constant. His velocity after five second will be equal to his initial velocity.#keeplearning dude:)A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the
end the compass points away from as "S" (south), and the other end that the compass points toward as "N" (north).
Record these poles in Figure 1.
Continue
Intro
Answer:
the red pointer on the magnet ( grey region) : points towards north
red pointer outside the magnet ( white region) is pointing towards south
Explanation:
please see the attached image
how can we know that atmosphere exert pressure explain with figure
Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.
The average temperature of the air in the room at 20 min is 23°C.
What is temperature?Temperature is the degree of hotness or coldness of the object.
A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.
The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.
Thus, the average temperature of the air in the room at 20 min is 23°C.
Learn more about temperature.
https://brainly.com/question/11464844
#SPJ2
g A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0kg and diameter 78.0cm. The power is off for 34.0s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.At what rate is the flywheel spinning when the power comes back on?
Answer:
[tex]10.54\ \text{rad/s}[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
[tex]\omega_f[/tex] = Final angular velocity
t = Time = 34 s
[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]
[tex]\alpha[/tex] = Angulr acceleration
From the kinematic equations of angular motion we have
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]
The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].
In an elastic collision between a moving 10-kg mass and a stationary 10-kg mass half the momentum is transferred to the stationary mass. In this situation the total kinetic energy after the collision is less than it was before the collision. Where did the kinetic energy go?
A) The kinetic energy was destroyed during the collision.
B) Some of the kinetic energy was turned into momentum during the collision.
C) Some of the kinetic energy was turned into heat or used to deform the masses.
D) Some of the kinetic energy was turned into potential energy during the collision.
Answer: C
Explanation:
USAtestprep
1. A group of students were trying to find the greatest
rebounded height of a rubber ball dropped on a basketball
court. They dropped from 3 different heights. The chart
below has their data.
26 cm
Drop Height Chart
Trials Drop height Rebound height
Trial 12 meters 103 cm
Trial 2% meter
Trial 31 meter 58 cm
Which explanation is the best reason for why trial 1 has the
greatest rebound height?
A. The speed of the ball is determined by the distance it
travels.
B. The force applied to the ball is a balanced force.
C) The greater the force applied to the ball the greater the
change in motion.
D. The closer the ball is to the ground the more gravity it
has.
Answer:
D th
Explanation:
D B. The force applied to the ball is a balanced force.
Why don’t the northern and Southern Hemisphere experience summer at the same time?
Answer:
It is because of the tilt of the earth.
Explanation:
the earth is tilted at 23.5 degrees. this makes it so that either the northern or southern hemisphere will be exposed to more rays from the sun. In the areas that are getting more rays from the sun, it gets warmer. Think about it like this, because the earth is tilted, part of it is more in the shade and part of it is more in the light. And its colder in the shade, so thats why seasons happen and why they dont happen at the same time.
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
You and a friend are playing with a Coke can that you froze so it's solid to demonstrate some ideas of Rotational Physics. First, though, you want to calculate the Rotational Kinetic Energy of the can as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. [The freezing only matters because if there is liquid inside, the calculation for the Moment of inertia becomes more complicated]. A Coke can can be modeled as a solid cylinder rotating about its axis through the center of the cylinder. This can has a mass of 0.33 kg and a radius of 3.20 cm. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with a linear velocity of 6.00 meters / second in the counter-clockwise (or positive) direction. You can use this to determine the angular velocity of the can (since it is rolling without slipping). What is the Total Kinetic Energy of the Coke can
Answer:
K_{total} = 8.91 J
Explanation:
In this exercise you are asked to find the kinetic energy of the can of coca-cola
K_total = K_ {Translation} + K_ {rotation}
the translational kinetic energy is
K_ {translation} = ½ m v²
the kinetic energy of rotation is
K_ {rotation} = ½ I w²
The moment of inertia of a cylinder is
I = ½ m r²
we substitute
K_ {total} = ½ m v² + ½ (½ m r²) w²
angular and linear velocity are related
v = w r
we substitute
K_ {total} = ½ m v² + ¼ m r² v² / r²
K_ {total} = m v² (½ + ¼)
K_ {total} = ¾ m v²
let's calculate
K_ {total} = ¾ 0.33 6.00²
K_{total} = 8.91 J
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
A 0.5kg ball of clay originally moving at 6 m/s strikes a wall and comes to rest in 0.25s, what is the magnitude of the impulse given to the ball of clay?
A) 0.75 kg m/s
B) 1.5 kg m/s
C) 3.0 kg m/s
D) 12 kg m/s
Answer:
C I did USA testprep
Explanation:
What is surface tension
Answer:
Surface tension is, the surface where the water meets the air, water molecules cling even more tightly to each other.
1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
What is not the ideal location for a radioactive waste storage facility?
A.
in an area that already has a lot of background radiation
B.
in an area that has few earthquakes
C.
far away from ground water
D.
in an area that is unpopulated with people
Answer:
D
Explanation:
Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 132 W/m2 . You may want to review (Pages 897 - 898) . Part A What is the angle from vertical of the axis of the second polarizing filter
Answer:
θ = 32.4º
Explanation:
For this exercise let's use Malus's law
I = Io cos² θ
in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical
I₀ = 370/2 = 185 W / m²
this is the radiation that affects the second polarizer, let's apply the expression of Maluz
θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])
θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])
θ = cos⁻¹ (0.844697)
θ = 32.4º
1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20
Answer:
It will go up to 93.75 m before it is moving at 20 m/s
Explanation:
As we know that
[tex]v^2 - u^2 = 2aS[/tex]
here v is the final speed i.e 20 m/s
u is the initial speed i.e 5 m/s
a is the acceleration due to gravity i.e 2 m/s^2
Substituting the given values in above equation, we get -
[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters
A truck is traveling on a level road. The driver suddenly applies the brakes, causing the truck to decelerate by an amount g/2. This causes a box in the rear of the truck to slide forward. If the coefficient of sliding friction between the box and the truckbed is 2/5, find the acceleration of the box relative to the truck and relative to the road.
Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity or the boy skater
his mass is 50 kg?
Question: Two people stand facing each other at a roller-skating rink then push off each other. If the girl skater has a mass of 30 kg and moves backward at 5 m/s, what is the velocity of the boy skater if his mass is 50 kg?
Answer:
3 m/s
Explanation:
Applying,
The Law of conservation of momentum
Momentum of the girl skater = momentum of the boy skater
MV = mv...................... Equation 1
Where M = mass of the girl skater, V = velocity of the girl skater, m = mass of the boy skater, v = velocity of the boy skater
From the question, we were asked to calculate v
v = MV/m.................. Equation 1
Given: M = 30 kg, V = 5 m/s, m = 50 kg
Substitute these values into equation 1
v = (30×5)/50
v = 3 m/s
Hence the velocity of the the boy skater is 3m/s
physics grade9 teacher guide
Answer:
huh
Explanation:
If F = force, which equation illustrates the Law of Conservation of Momentum?
A) F1 = F2
B) F1 = - F2
C) - F1 = -F 2
D) F1 + - F2 = F3
Answer:
b
Explanation:
f1=-f2 that could be thank u
4. Speedy leaves the ground with an initial vertical velocity of 53 m/s and a horizontal velocity of 42 m/s.
How much time does he spend in the air?
How far (horizontally) does he travel during this time?
5. The Angry Bird is fired at an angle of 35 above the horizontal at a speed of 72 m/s.
Draw the initial velocity vector
Determine the initial horizontal velocity
Determine the initial vertical velocity
How much time does it spend in the air?
What horizontal distance does it go?
what kind of charge does an object have if it has extra positive charges
Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is
Answer:
The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].
Explanation:
The work done is given as in terms of
[tex]W=\Delta TE[/tex]
Where ΔTE is the change in total energy.
This is given as
[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]
Rearranging it in terms of K_x gives
[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.07 kg. They then hang the object on a pivot located 0.155 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 247 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
I = 0.65 kgm²
Explanation:
Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.
T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.
Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation
So, T = 2.186 s
We now find I by making it subject of the formula in the equation for T.
So,
T = 2π√(I/mgh)
dividing both sides by 2π, we have
T/2π = √(I/mgh)
squaring both sides, we have
(T/2π)² = [√(I/mgh)]²
T²/4π² = I/mgh
multiplying both sides by mgh, we have
T²mgh/4π² = I
I = T²mgh/4π²
substituting the values of the variables into the equation, we have
I = T²mgh/4π²
I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 29.539 kgm²/4π²
I = 0.748 kgm²
Now I = I' + mh² (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.
So, I' = I - mh²
Substituting the values of the variables into the equation, we have
I' = I - mh²
I' = 0.748 kgm² - 4.07 kg × (0.155 m)²
I' = 0.748 kgm² - 4.07 kg × 0.02403 m²
I' = 0.748 kgm² - 0.098 kgm²
I = 0.65 kgm²
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.
Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
Determine the Thrust developed
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : calculate the area of the duct
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
next : calculate the velocity of propeller
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
Finally determine the thrust developed
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft. You begin to run from the inside to the outside along a radius. Your peak velocity with respect to the carousel is 6 mph and occurs at a radius of 8 ft.What are your maximum Coriolis acceleration magnitude and its directionwith respect to the carousel
Answer:
magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
Explanation:
Given the data in the question;
Speed of carousel N = 24 rpm
From the diagram below, selected path direction defines the Axis of slip.
Hence, The Coriolis is acting along the axis of transmission
Now, we determine the angular speed ω of the carousel.
ω = 2πN / 60
we substitute in the value of N
ω = (2π × 24) / 60
ω = 2.5133 rad/s
Next, we convert the given velocity from mph to ft/s
we know that; 1 mph = 1.4667 ft/s
so
[tex]V_{slip[/tex] = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s
Now, we determine the magnitude of the Coriolis acceleration
[tex]a_c[/tex] = 2( [tex]V_{slip[/tex] × ω )
we substitute
[tex]a_c[/tex] = 2( 8.8002 ft/s × 2.5133 rad/s )
[tex]a_c[/tex] = 44.235 ft/s²
Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission
6. If an object accelerates at 3m/s/s, how long does it take for the object to travel at a speed of 12 m/s.
Answer:
4 seconds
Explanation:
Assuming that the object started from rest,
v = at
--> t = v/a = (12 m/s) / (3 m/s^2)
= 4 seconds
