Answer:
Stellar disk: 185 ± 15 kly (kilo Langley/year)
Dark matter halo: 1.9 ± 0.4 Mly (mega light year) (580 ± 120 kpc)
Explanation:
The Milky Way galaxy contains the sun and the planets orbiting around it. This galaxy also contains up to 400 billion stars. The stellar disk of this galaxy is between 170,000 to 200,000 light years which is equal to a diameter of 52 - 61 kiloparsec.
Its thickness is measured at 0.3kpc which is also equal to 1000 light years. Its mass is 890 billion to 1.54 trillion times the mass of the sun. Its size is second to that of the Andromeda galaxy.
6. Thomas sits on a small rug on a polished wooden floor. The coefficient to kinetic friction between the rug and
the slippery wooden floor is only 0.12. If Thomas weighs 650 N, what horizontal force is needed to pull the rug
and Thomas across the floor at a constant speed?[78N]
1
Answer:
F=78 N
Explanation:
Taking to Thomas and the rug as a single system, if they slide across the floor at constant speed, this means that their acceleration is just zero.According to Newton's 2nd Law, if the acceleration is zero, this means that the net force applied is zero too.In the horizontal direction, there are two forces acting on Thomas and the rug (as a single system), the applied force, and the kinetic friction force, which must be equal and opposite each other:[tex]F_{app} = F_{kfr} (1)[/tex]
By definition, as the friction force is the horizontal component of the contact force, it can be expressed as follows:[tex]F_{1kfr} = \mu_{k} * F_{n} (2)[/tex]
where μk = coefficient of kinetic friction = 0.12
Fn = normal force
In this case, as the system boy+rug is not accelerating in the vertical direction, and the surface is level, the normal force (which is always perpendicular to the surface), must be equal to the force of gravity.Assuming that the mass of the rug is neglectable, we can write:[tex]F_{n} = F_{g} = m*g = 650 N (3)[/tex]
Replacing (3) and μk in (2)[tex]F_{1kfr} = \mu_{k} * F_{n} = 0.12 * 650 N = 78 N (4)[/tex]
From (1), we finally get:[tex]F_{app} = F_{kfr} = 78 N (5)[/tex]
Tripling the wavelength of the radiation from a monochromatic source will change the energy content of the individually radiated photons by what factor
Answer: By a factor of 1/3.
Explanation:
For a photon with wavelength λ, the energy is written as:
E = h*c/λ
where:
h is the Planck's constant:
h = 6.63*10^(-34) Js
c is the speed of light:
c = 3*10^8 m/s
Now, if we triple the wavelength of this photon, then the energy will be:
E' = (h*c)/(3*λ)
We rewrite this as:
E' = (1/3)*(h*c/λ)
And (h*c/λ) was the previous energy:
(h*c/λ) = E
Then we can replace that in the above equation to get:
E' = (1/3)*(h*c/λ) = (1/3)*E
Then if we triple the wavelength, it will change the energy content of the individually radiated photons by a factor of 1/3.
In October 1997, Andy Green broke the sound barrier on land in a jet-powered car. Green's car had accelerated from rest to 763 mi/h (341 m/s) over a 5.0-mi (8047-m) stretch. Assuming constant acceleration, over what time interval had he reached the top speed
Answer: the time interval Green had to reached the top speed is 47.2 secs
Explanation:
Given the data in the question;
first we determine Green's acceleration.
from third equation of motion, v² - u² = 2as
a = v² - u² / 2s
we substitute,
a = ( (341 m/s)² - (0)² ) / 2(8047)
a = (116281 - 0) / 16094
a = 7.23 m/s²
Now we determine the time interval need to reach top or maximum speed,
v = u + at
at = v - u
t = (v - u) / a
so we substitute
t = (341 - 0) / 7.23
t = 341 / 7.23
t = 47.2 secs
Therefore, the time interval Green had to reached the top speed is 47.2 secs
An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length. The distance between the lenses is 5.0 cm. Find the location and size of the final image.
Answer:
122.735 behind converging lens ; 2.16
Explanation:
Given tgat:
Object distance, u = 29 cm
Image distance, v =
Focal length, f = - 19 (diverging lens)
Mirror formula :
1/u + 1/v = 1/f
1/29 + 1/v = - 1/19
1/v = - 1/19 - 1/29
1/v = −0.087114
v = −11.47916
v = -11.48
Second lens
Object distance :
u = 11.48 + 11 = 22.48 cm
1/v = 1/19 - 1/22.48
1/v = 0.0081475
v = 1 / 0.0081475
v = 122.735 cm
122.735 behind second lens
Magnification, m
m = m1 * m2
m = - v / u
Lens1 :
m1 = -11.48 / 29 = - 0.3958620
m2 = - 122.735 / 22.48 = - 5.4597419
Hence,
- 0.3958620 * - 5.4597419 = 2.16
When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of the square. Suppose three identical charges of q are placed on the remaining three corners of the square (so that each corner has q). What is the magnitude of the net electric field at the center of the square
Answer:
0 N/C
Explanation:
Since the electric field due toeach charge q is at the same distance from the center of the square and has a magnitude E = 2 N/C, we resolve the components of each electric field to vertical and horizontal.
So, for each charge, the magnitude of the horizontal component is E' = Ecos45 and the magnitude of the vertical component is E" = Esin45.
For the charges on the bottom half of the square, the direction of the vertical component is upwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the bottom two charges is 2Esin45.
For the charges at the upper half of the square, the direction of the vertical component is downwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the top two charges is -2Esin45.
So, the resultant electric field at the center of the square is thus 2Esin45 + (-2Esin45) = 2Esin45 - 2Esin45 = 0 N/C
A 235 kg object and a 1.37×1012 kg are located 2.59×104 m away from each other. What is the force due to gravity between the two objects?
Answer:
F = 3.2 x 10⁻⁵ N
Explanation:
The gravitational force of attraction between the two objects is given by Newton's Gravitational law through the following formula:
[tex]F = \frac{Gm_{1}m_{2}}{r^{2}}[/tex]
where,
F = gravitational force = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²
m₁ = mass of object 1 = 235 kg
m₂ = mass of object 2 = 1.37 x 10¹² kg
r = distance between objects = 2.59 x 10⁴ m
Therefore,
[tex]F = \frac{(6.67\ x\ 10^{-11}\ Nm^{2}/kg^{2})(235\ kg)(1.37\ x\ 10^{12}\ kg)}{(2.59\ x\ 10^{4}\ m)^{2}}[/tex]
F = 3.2 x 10⁻⁵ N
Answer:
the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.
Explanation:
Given;
mass of the first object, m₁ = 235 kg
mass of the second object, m₂ = 1.37 x 10¹² kg
distance between the two object, r = 2.59 x 10⁴ m
The gravitational force between the two object is calculated as;
[tex]F= \frac{Gm_1m_2}{r^2}[/tex]
where;
G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²
[tex]F= \frac{(6.67\times 10^{-11})(235)(1.37\times 10^{12})}{(2.59\times 10^4)^2} \\\\F = 3.2 \times 10^{-5} \ N[/tex]
Therefore, the force due to gravity between the two objects is 3.2 x 10⁻⁵ N.
What is a Wave In your own words
A hunter points a rifle directly at a coconut that he intends to shoot off a tree. It so happens that the coconut falls from the tree at the exact instant the hunter pulls the trigger. What happens to the bullet?
Answer:
Explanation:
The bullet will hit the coconut .
Both coconut and bullet will fall under the gravitational pull of the earth . They will experience same acceleration because acceleration does not depend upon mass . So both bullet and coconut will fall by the same distance vertically .
So bullet will exactly hit the coconut .
How much power is used if an athlete lifts a 50N weight up 2 meters in 10 seconds?
Answer:
the power used by the person is 10 W.
Explanation:
Given;
weight lifted, W = 50 N
distance through which the weight was lifted, d = 2 m
time of motion, t = 10 s
The power used by the person is calculated as;
[tex]P= FV\\\\P = F \ \times \ \frac{d}{t} \\\\P = 50 \ \times \ \frac{2}{10} \\\\P = 10 \ W[/tex]
Therefore, the power used by the person is 10 W.
A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field at distance r from the axis inside the cylinder in terms of r/R.
Answer:
[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]
Explanation:
Given that:
[tex]\rho_o = (10^{-3} ) \ c/m^3[/tex]
R = (0.1) m
To find the electric field for r < R by using Gauss Law
[tex]{\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)[/tex]
For r < R
[tex]Q_{enclosed}=(\rho) ( \pi r^2 ) l[/tex]
[tex]E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}[/tex]
[tex]E= \dfrac{\rho ( r)}{2 \varepsilon_o}[/tex]
where;
[tex]\varepsilon_o = 8.85 \times 10^{-12}[/tex]
[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]
[tex]E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}[/tex]
[tex]E = (0.56 \times 10^8 ) r \ \ N/c[/tex]
A pair of in-phase stereo speakers is placed side by side, 1.26 m apart. You stand directly in front of one of the speakers, 2.79 m from the speaker. What is the lowest frequency that will produce destructive interference at your location? Group of answer choices
Answer:
1264 Hz
Explanation:
We are given;
Your distance to the front of first speaker: L1 = 2.79 m
Distance between speakers; d = 1.26m
Now, your distance to second speaker can be calculated using pythagoras theorem.
Thus;
L2 = √(1.26² + 2.79²)
L2 = 3.0613 m
Let's find the wavelength from the formula;
(L1 - L2) = nλ
Where n is the order and could be; n = 0,1,2,3...
Making λ which is the wavelength the subject, we have;
(L1 - L2)/n = λ
The least frequency will occur at the wavelength when n = 1.
λ = (2.79 - 3.0613)/1
λ = -0.2713 m
We will take the absolute value and thus;
λ = 0.2713 m
Frequency is gotten from;
f = v/λ
Where v is speed of sound = 343 m/s
Thus;
f = 343/0.2713
f ≈ 1264 Hz
The energy of an electromagnetic wave changes proportionally to which other
property?
A. Frequency
B.Speed
C.Shift
D.Wavelength
I know it’s not wavelength
Answer: A. frequency
Explanation:
If it’s not wavelength then it has to be frequency
The acceleration due to gravity on or near the surface of Earth is 32 ft./s/s. Neglecting friction, from what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft./s
Given :
The acceleration due to gravity on or near the surface of Earth is 32 ft/s/s
To Find :
From what height must a stone be dropped on Earth to strike the ground with a velocity of 136 ft/s.
Solution :
Initial velocity of stone, u = 0 ft/s.
Now, by equation of motion :
[tex]2as = v^2 -u^2 \\\\2\times 32 \times s = 136^2 -0^2\\\\s = \dfrac{136^2}{2\times 32}\ ft\\\\s = 289 \ ft[/tex]
Therefore, height from which stone is thrown is 289 ft.
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the electrical force that each charge exerts on the other one
Answer:
Remains the same
Explanation:
Determine the values of mm and nn when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,000,000,000 kgkg.
Answer:
"m" and "n" are 5.97 and 24 respectively.
Explanation:
Standard form which is a scientific notation [m × 10^n] can be regarded as way to reduce large figures to small one in a decimal firm and this is usually done for conviniency sake.
✓Let us find "m" , to do this we will shorten the long number so that it will be from 1-9. And this is 5.97, hence
m = 5.97
✓ " n" can be determined by counting the digits from our right hand then stop where the decimal point was put when we were finding our "m". Hence
n= 24
✓ if we input the values to the scientific expresion above, we have
m = 5.97 and n = 24
Hence, the Standard form = 5.97 × 10^24 kg which is the scientific notation.
The net electric flux through a cubic box with sides that are 22.0 cm long is 4950 N⋅m2/C . What charge is enclosed by the box?
Answer:
[tex]Q_{net}=4.38\,\,10^{-8}\,C[/tex]
Explanation:
We use Gauss's Law for the flux over a closed surface equal the net charge inside the closed surface divided the permitivity of space [tex]\epsilon_0=8.85\,\,10^{-12}\,\,\frac{C^2}{N*m^2}[/tex]
Therefore, by knowing the flux, we can estimate the net charge inside the cubic box with the product:
[tex]\Phi=\frac{Q_{net}}{\epsilon_0} \\Q_{net}=4950\,*\,8.85\,\,10^{-12} \,\,C\\Q_{net}=4.38\,\,10^{-8}\,C[/tex]
if we are making koolaid with sugar, koolaid powder and water whitch part is the solvent
Answer:The powder of Kool Aid crystals are the solute. The water is the solvent and the delicious Kool Aid is the solution.... .-.
Explanation:
5. Candance got $0.22 in change when she bought chips at the
store. What percent of a dollar does her change represent?
Answer:
22%
Explanation:
Given parameters:
The amount of change = $0.22
Unknown
What percentage of a dollar the change represents = ?
Solution:
A dollar is made up of 100cents;
So;
$0.22 represents 22cents
Percentage of dollar this represents = [tex]\frac{22}{100}[/tex] x 100 = 22%
If an object is thrown downward at 4.73 m/s and it falls for 6.21 seconds before landing, how fast is it falling the instant before it lands?
Answer:
65.59 m /s
Explanation:
Initial velocity u = 4.73 m/s
Final velocity v = ?
time t = 6.21 s
acceleration = g = 9.8 m /s
v = u + gt
= 4.73 + 9.8 x 6.21
= 65.59 m /s .
An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat curve with radius of 65.0 m be rated
Answer:
The maximum speed is 21.39 m/s.
Explanation:
Given;
radius of the flat curve, r₁ = 150 m
maximum speed, [tex]v_{max}[/tex] = 32.5 m/s
The maximum acceleration on the unbanked curve is calculated as;
[tex]a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2[/tex]
the radius of the second flat curve, r₂ = 65.0 m
the maximum speed this unbanked curve should be rated is calculated as;
[tex]a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s[/tex]
Therefore, the maximum speed is 21.39 m/s.
On a day when the wind is blowing toward the south at 4 m/s, a runner jogs east at 5 m/s. What is the velocity (speed and direction) of the air relative to the runner
Answer:
[tex]v=5m/s[/tex]
Direction:SW(south-west)
Explanation:
From the question we are told that
Velocity and direction of Runner [tex]V_r= 5m/s ,East[/tex]
Velocity and direction of Air[tex]V_a= 4m/s ,South[/tex]
Generally the the resultant velocity v is mathematically given as
[tex]v=\sqrt{V_a^2+V_r^2}[/tex]
[tex]v=\sqrt{4^2+5^2}[/tex]
[tex]v=5m/s[/tex]
The resultant velocity is towards the south-west
Which do you suppose exerts more pressure on the ground—an elephant or a lady standing on spike heels? (Which will be more likely to make dents in a linoleum floor?) Can you approximate a rough calculation for each?
Answer:
A woman with a spike heel
Explanation:
Since we're assuming here, we proceed to assuming that the mass of the elephant is 6000 kg, and the mass of the lady is 60 kg. Then,
a woman with a spike heel will exert more pressure on the ground than an elephant weighing 6000 kg. This is because, the force exerted by the woman is a 588 N and she has a, say 1 cm² spike heel. This puts half her weight on each foot, and if evenly distributed to half on her heel and half on her sole, the pressure being exerted by each heel is 147 N/ 2cm² = 147 N/cm². On the other hand, if a 58,860 N elephant with an 1000 cm² feet is exerting 1/4 its weight on each foo. Then it has 14715 N/1000 cm² = 14.7 N/cm²; which is exactly 10 times lesser pressure than what the woman is exerting. So, the woman with the hell exerts more pressure, than the elephant will.
7. A girl pushes her little brother on his sled with a force of 300. N for 750. m. How much work is this if the force of friction acting on the sled is (a) 200. N. (b) 300. N?
Answer:
a) 75000Joules
b) 0Joules
Explanation:
Workdone = Force * Distance
Given
distance= 750m
Force = 300N
a) If the frictional force = 200N
The Total force = 300N - 200N = 100N
Work done = 100 * 750
Workdone = 75,000Joules
Hence the workdone if the force of friction is 200N is 75,000Joules
b) If the frictional force = 300N
The Total force = 300N - 300N = 0N
Work done = 0* 750
Workdone = 0Joules
Hence the workdone if the force of friction is 300N is 0Joules i.e no work will be done on the sled
A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.10 103 N/C, while the magnetic field is 0.360 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.20 cm. Find the charge-to-mass ratio of the particle.
Answer:
The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg
Explanation:
From the formulae
F = qvB and F = mv²/r
Where F is Force
q is charge
v is speed
B is magnetic field strength
m is mass
and r is radius
Then,
qvB = mv²/r
qB = mv/r
We can write that
q/m = v/rB ---- (1)
Also
From Electric force formula
F = Eq
Where E is the electric field
and magnetic force formula
F = Bqv
Since, electric force = magnetic force
Then, Eq = Bqv
E = Bv
∴ v = E/B
Substitute v = E/B into equation (1)
q/m = (E/B)/rB
∴ q/m = E/rB²
(NOTE: q/m is the charge to mass ratio)
From the question,
E = 3.10 ×10³ N/C
r = 4.20 cm = 0.0420 m
B = 0.360 T
Hence,
q/m = 3.10 ×10³ / 0.0420 × (0.360)²
q/m = 569517.9306 C/kg
q/m = 5.7 × 10⁵ C/kg
Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.
Sound waves can be modeled by the equation of the form y=20sin(3t+theta). Determine what type of interference results when sound waves modeled by the equations y=20sin(3t+90) and y=20sin(3t+270) are combined.
Answer:
go to the link quizzlet it will give you tha answer
Explanation:
An engine has an output energy of 2,400 J in 10 seconds. What is its average power in watts?
Answer:
P = 240 W
Explanation:
By definition, the average power is the rate of change of Energy (in Joules) regarding time (in seconds), as follows:[tex]P = \frac{\Delta E}{\Delta t} (1)[/tex]
Replacing in (1) by the givens of ΔE = 2, 400 J and Δt = 10 sec, we can find the average power in Watts as follows:[tex]P = \frac{\Delta E}{\Delta t} =\frac{2,400J}{10s} = 240 W (2)[/tex]
A 2.1-F capacitor is fully charged by a 6.0-V battery. The battery is then disconnected. The capacitor is not ideal and the charge slowly leaks out from the plates. The next day, the capacitor has lost half its stored energy. Calculate the amount of charge lost.
Answer:
6.3 C
Explanation:
From the question,
Energy lost by the capacitor(half it stored energy) = 1/4CV².......................... Equation 1
Where C = Capacitance of the capacitor, V = Volatge across the parallel plate of the capacitor.
E = 1/4CV²
Given: C = 2.1 F, V = 6.0 V
Substitute these values into equation 1
E = 1/4(2.1)(6²)
E = 18.9 J
But,
E = 1/2QV..................... Equation 2
Where Q = amount of charge lost
Make Q the subject of the equation
Q = 2E/V................. Equation 3
Give: E = 18.9 J, V = 6.0 V
Q = 2(18.9)/6
Q = 6.3 C.
At a sports car rally, a car starting from rest accelerates uniformly at a rate of 5 m/s/s over a straight-line distance of 291 m. How long (in seconds) did it take the car to travel the 291 m
Answer:
10.8 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Acceleration (a) = 5 m/s/s
Distance travelled (s) = 291 m
Time (t) taken =?
We can calculate the time taken for the car to cover the distance as follow:
s = ut + ½at²
291 = 0 × t + ½ × 5 × t²
291 = 0 + 2.5 × t²
291 = 2.5 × t²
Divide both side by 2.5
t² = 291 / 2.5
t² = 116.4
Take the square root of both side
t = √116.4
t = 10.8 s
Thus, it will take the car 10.8 s to cover the distance.
The time taken by the car to travel 291 m is calculated using the kinematics equation and it is obtained to be 10.8 seconds.
Motion in a Straight LineHere, it is given that;
The acceleration of the car is,
[tex]a = 5\,m/s^2[/tex]
The distance covered by the car is,
[tex]s = 291\,m[/tex]
Also, the car starts from rest, that means;
[tex]u = 0\,m/s[/tex]
We can use the second kinematics equation to find the time taken by the car to travel the given distance.
[tex]s = ut + \frac{1}{2}\,at^2\\291\,m = (0\,m/s \times t)+ \frac{1}{2} (5\,m/s^2\times t^2)\\\\\\291\,m = (2.5\,m/s^2\times t^2)[/tex]
implies
[tex]t = \sqrt{\frac{291\,m}{2.5\,m/s^2}}=10.78\,s\approx10.8\,s[/tex]
Learn more about 'motion in a straight line' here: https://brainly.com/question/19558242
1. Which of the following will be surrounded by a magnetic field? *
1Two copper wires held close together.
2A pot being heated on an electric stove.
3A wire that is conducting an electrical current.
4An aluminum rod that has been touched by a magnet.
Answer:
4th answer
Explanation:
A wooden block is let go from a height of 5.80 m. What is the velocity of the block just before it hits the ground?
Given :
A wooden block is let go from a height of 5.80 m.
To Find :
The velocity of the block just before it hits the ground.
Solution :
We know, by equation of motion :
[tex]v^2 - u^2 = 2as[/tex]
Here, a = g = 9.8 m/s²( Acceleration due to gravity )
Putting all given values in above equation, we get :
[tex]v^2 - u^2 = 2as\\\\v^2 -0 = 2\times 9.8 \times 5.8 \\\\v = \sqrt{2\times 9.8 \times 5.8 } \ m/s\\\\v = 10.66\ m/s[/tex]
Hence, this is the required solution.
