Answer:
Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.
Explanation:
Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.
Let's model the Milky Way galaxy as a disk.
The volume of a disk is:
V
=
π
⋅
r
2
⋅
h
Plugging in our numbers (and assuming that
π
≈
3
)
V
=
π
⋅
(
10
21
m
)
2
⋅
(
10
19
m
)
V
=
3
×
10
61
m
3
Is the approximate volume of the Milky Way.
Now, all we need to do is find how many stars per cubic meter (
ρ
) are in the Milky Way and we can find the total number of stars.
Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of
4
×
10
16
m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.
ρ
=
n
V
Using the volume of a sphere
V
=
4
3
π
r
3
ρ
=
1
4
3
π
(
4
×
10
16
m
)
3
ρ
=
1
256
10
−
48
stars /
m
3
Going back to the density equation:
ρ
=
n
V
n
=
ρ
V
Plugging in the density of the solar neighborhood and the volume of the Milky Way:
n
=
(
1
256
10
−
48
m
−
3
)
⋅
(
3
×
10
61
m
3
)
n
=
3
256
10
13
n
=
1
×
10
11
stars (or 100 billion stars)
Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.
Answer:
Mercury, 46,001,272 km from the sun at the nearest point.
Explanation:
Two point charges of equal magnitude are 7.5cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 49N/CFind the magnitude of the charges (pC)
Answer:
q 1 = q 2 = Q
E = k * Q / r²
The combined electric field:
E = 2 * k * Q / r²
k = 9 · 10^9 Nm²/C²
r = 0.075 : 2 = 0.0375 m
45 N/C = 2 · 9 · 10^9 · Q / 0.0375²
45 = 18 · 10 ^9 Q / 0.0014062
Q = 45 · 0.0014062 / 18 · 10^9
Q = 0.003515 · 10^(-9) C
Q = 3.5155 · 10 ^ (-12) C = 3.5155 pC
calculate horizontal distance travelled by a ball travelling with a speed of 20root2 mper sec without hitting ceiling of height 20 m per sec
Calculate the horizontal distance travelled by a ball throw with a velcoity 20 sqrt 2 ms^(-1) without hitting the ceiling of an anditorium of heith 20 m. Use g= 10 ms^(-2). =(20√2)2sin2×45∘10=9800×1)10=80m .
HOPE SO IT HELPS YOU
What is the author's MAIN purpose for including information about viruses? D to suggest that it could be dangerous for explorers to bring home life from other planets B to provide an example that illustrates the difficulty of defining life in some cases on Earth С to demonstrate the need for extra protective gear when explorers go into space D to highlight the differences between natural life and artificial life mixed in a lab Back Submit
Answer:
B
Explanation:
A race car goes from a complete stop at the start line to 150 miles per hour in 5 seconds. What is its acceleration? Show your work.
Answer:
Explanation:
150/5 = 30
30mph per 1 second
if you jog at a speed of 1.5m/s for 20 seconds how far di you travel
Answer: 30m
Explanation:
Given:
Speed: 1.5m/s
Time: 20 seconds
Distance = speed × time
Distance = 1.5 × 20
= 30m
Therefore you will travel 30m
Must click thanks and mark brainliest
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?
Answer:
A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?
Explanation:
Please help wil give brainiest & 40p.
Answer:
This is the answer
Explanation:
You can find the ans in the photo I attached.
The car has a mass of 0·50 kg. The boy
now increases the speed of the car to 6·0
ms-1 . The total radial friction between
the car and the track has a maximum
value of 7.0 N. Show by calculation that
the car cannot continue to travel in the circular path.
Answer:
A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
What is meant by centripetal force ?Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.
Here,
Mass of the car, m = 0.5 kg
Velocity of the car, v = 6 m/s
Radial friction between the car and the track, f = 7 N
The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.
So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.
So,
mv²/r > f
0.5 x 6²/r > 7
Therefore, the radius of the circular track,
r < 18/7
r < 2.57 m
Hence,
The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.
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. Two charges, Q1 and Q2 , are separated by a certain distance R. If the magnitudes of the charges are halved, and their separation is also halved, then what happens to the electrical force between these charges
Answer: Magnitude of electrical force stays the same.
Explanation:
Equation:
[tex]F_{e} =k\frac{Q_{1}Qx_{2} }{r^{2} }[/tex]
Since the magnitude of each charge is halved.
and
the separation is halved.
[tex]F_{e} =k\frac{(.5Q_{1}*.5Q_{2} }{(.5r)^{2} }[/tex]
[tex]F_{e} =k\frac{.25*Q_{1}Qx_{2} }{.25*r^{2} }[/tex]
Cancel out .25 on the numerator and denominator. Leaving the original equation.
The ratio of RMS of velocity , Probable Velocity and Average velocity is ??
Answer:
2 :π8 :3
The ratio between the most probable velocity,mean velocity and root mean square velocity is 2 :π8 :3. Vrms=M3RT.
Explanation:
pls mark brainliest
The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?
Answer:
2.5 km
Explanation:
Answer:
2.5 km
Explanation:
Distance = speed x time
So =5 x 0.5
a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
Please help! Can give brainliest too.
Explanation:
[tex]v = \sqrt{2ax}[/tex]
Take the square of both sides:
[tex]v^2 = \left(\sqrt{2ax}\right)^2 = 2ax[/tex]
Divide both sides by 2a and you will get
[tex]x = \dfrac{v^2}{2a}[/tex]
who was the youngest female to give birth
Answer:
5 years old 7 months and 21 days
Explanation:
Answer:
lina marcela medina de jurado
Can you please help me please?
Explanation:
CH3CH2OH
That is the answer I hope this helps
An electric motor and a generator are similar in that both
A. Produce mechanical energy.
B. Use electricity.
C. Transform energy into a different form.
D. Create energy.
C. transform energy into a different from.
I hope this helps
(a) Joseph was pushing a car.On his observation found that it was easier to keep the car in moti with a smaller force than to start its motion. Explain why?
Answer:
Newton's laws states that 'what is in motion, stays in motion until a external force is applied', when a car is already moving, it's in motion already but to start a car, you need to apply some force to get it going, that force isn't required for a moving car.
what happens to the weight of the body when it is falling freely under the action of gravity
Answer:
A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object
Explanation:
plz mark me as brainliest
Answer:
Gradually increases until the maximum weight reaches the surface of the earth
Explanation:
How many different ingredients will you need…
Answer:
11
Explanation:
hope this helps!!!!!
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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8. A mass of 10 Kg is accelerating at 3 m/s2. What is the applied net force?
Answer:
Explanation:
F = ma
F = (10)(3)
F = 30 N
Answer:
[tex]\boxed {\boxed {\sf 30 \ Newtons}}[/tex]
Explanation:
We are asked to find the applied net force. According to Newton's Second of Law, force is the product of mass and acceleration.
[tex]F= m \times a[/tex]
The object has a mass of 10 kilograms and it is accelerating at 3 meters per second squared.
m= 10 kg a= 3 m/s²Substitute the known values into the formula.
[tex]F= 10 \ kg \times 3 \ m/s ^2[/tex]
Multiply.
[tex]F= 30 \ kg \times m/s^2[/tex]
1 kilogram meter per second squared is equal to 1 Newton, so our answer of 30 kg × m/s² is equal to 30 N.
[tex]F= 30 \ N[/tex]
The applied net force is 30 Newtons.
what does displacement consists of
Answer:
The displacement of a body has two components: rigid-body displacement and deformation. A rigid-body displacement consists of a simultaneous translation and rotation of the body without changing its shape or size.
Explanation:
What is the momentum of a 36.9 N bowling ball with a velocity of 7.56 m/s?
Answer:
momentum (m)=36.9N
velocity (v)=7.56m/s
now,
momentum (m)=m×v
36.9=m×7.56
36.9÷7.56=m
m=4.89kg
Which units are being recalled?
—-
Answer:
a c e f
Explanation:
Start with the last year then the month. to eliminate wrong choices
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.
Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.
From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)
Where:
[tex]W_{T}[/tex] - Work done by tension, in joules.
[tex]m[/tex] - Mass of the sled-victim system, in kilograms.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.
[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.
If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]
[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]
[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]
[tex]W_{T} = 1662.544\,J[/tex]
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
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an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons
pls write the formula too
Element is
[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]
Atomic number=13Mass number=27[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]
And[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]
And
[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]
a standard bathroom scale is placed on an elevator. A 34 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.4 m/s2
Answer:F = 255 N
Explanation:
It is given that,
Mass of the boy, m = 25 kg
Acceleration of the elevator,
The elevator is accelerating in upward direction. The net force acting on the boy is given by :
g is the acceleration due to gravity
F = 255 N
The scale reading is 255 N as it begins to accelerate upward. hence, this is the required solution.
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
Where does a body have more weight the poor at the eqator of the earth.
Answer:
Explanation:
Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.
Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.
As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.
