Answer:
K_A = 32.2 10⁶ J
Explanation:
In this exercise we must relate the quantities given to find the kinetic energy
Asteroid A data
m_A = 3.5 m_B
v_A = 2.0 v
they also give the value of the kinetic energy of asteroid A
K_B = 2.3 10⁶ J
the expression for scientific energy is
K = ½ m v²
let's replace
K_A = ½ m_a V_a2
K_A = ½ 3.5 m_B (2.0 v_B)^2
K_A = 3.5 2² (½ m_B v_B²)
K_A = 14 K_B
K_A = 32.2 10⁶ J
A 0.10 kg point mass moves in a circular path with a radius of 0.36 m with a net force of 10.0 N toward the center of the circle. Select all of the following that are true statements.
a. The velocity of the object is 6 m/s toward the center of the circle.
b. The speed of the object is 6 m/s and decreasing.
c. The speed of the object is 6 m/s and increasing.
d. The velocity of the object is a constant 6 m/s.
e. The speed of the object is a constant 6 m/s.
Answer:
e. The speed of the object is a constant 6 m/s
Explanation:
Since the net force is towards the centre , hence there is no tangential acceleration . Only centripetal acceleration is there . Hence point mass is moving with uniform speed . Let it be u .
Centripetal force = m v² / r , r is radius of circular path .
Putting the given values
.10 x v² / .36 = 10
v = 6 m /s
For this study, the researcher is analyzing data by using __________ measures. A. experimental B. qualitative C. quantitative D. naturalistic
Answer:
B. qualitative
Explanation:
Since in the question it is mentioned that it takes the teacher opinion with respect to the student lunchroom that arranged the opinion in four classifications and according to that the report is also written
So this research represents the quality of the data collected as it does not specify the rating in terms of liking the student lunchroom
Therefore the correct option is B. qualitative
Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy U from the light. The linear momentum these objects will receive is such that:
Answer:
absorbent p = S / c
reflective p = 2S/c
Explanation:
The moment of radiation on a surface is
p = U / c
where U is the energy and c is the speed of light.
In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.
p = S / c
in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process
p = 2S/c
16. In single-slit diffraction, the central band gets thicker as the distance to the screen increases. True False
Answer:
the right answer is true
Answer:
True
Explanation:
A platinum wire is used to determine the melting point of indium. The resistance of the platinum wire is 2.000 Ω at 20°C and increases to 3.072 Ω as indium starts to melt. What is the melting point of indium? (The temperature coefficient of resistivity for platinum is 3.9 ×
Answer:
The melting point of indium is 157.436 degrees Celsius.
Explanation:
The resistance of the platinum wire, R1 = 2
The temperature at R1 is, T1 = 20 degrees Celsius.
The increased resistance, R2 = 3.072
Let the temperature at 3.072 = T2
Now find the temperature at which the indium starts melting.
We know that α = ( R2 - R1 ) / [ R1 × ( T2 - T1 ) ]
Given, α = 3.9 x 10^-3/ degrees Celsius.
T2- T1 = ( R2 - R1 ) / R1 α
T2 – T1 = (3.072 – 2) / (2 × 3.9 x 10^-3)
T2 – T1 = 137.436
T2 = T1 + 137.436
T2 = 20 + 137.436
T2 = 157.436 degree Celsius
Answer:
1,772 C
Explanation:
a box container both cube and five side pyramid the total number of objects is 17 the total number of side for the cube and pyramid is 95 how many cubes are in the box
Answer:
There are 10 cubes in the box
Explanation:
Let the total number of cubes be x and the total number of pyramids be y
Since there are 17 total objects;
Then;
x + y = 17 •••••••••(i)
Also, the total number of sides of cubes is 6 * x = 6x ( a single cube has 6 sides)
For the pyramid we have 5 * y = 5y
adding both gives the total number of sides
6x + 5y = 95 •••••• (ii)
From i, we cabs say y = 17-x
plug this in ii
6x + 5(17-x) = 95
6x + 85 -5x = 95
6x-5x = 95-85
x = 10
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad/s)t].
(a) What is the speed of the wave?
(b) What are the amplitudes of the electric and magnetic fields of this wave?
(c) What are the frequency, wavelength, and period of the wave? Is this light visible to humans?
Answer:
a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
Why does front side of spoon forms inverted image but the back side form opposite of inverted image?
Answer:
our face is outside the focal length of the concave side of the spoon. We see a virtual inverted image whereas in case of concave mirror we can see a virtual image which is erect.
What would you predict would occur if you were able to
place a rotating magnet near a coil of wire?
Will occur if you were able to place a rotating magnet near a coil of wire is continually rotating magnetic field would continuously induce current.
When is the operator faster approaching or moving away from the magnetic flux?
When the magnet moves close to the coil or rapidly increasing until the magnet is within the flux of the coil. As it passes through the coil, the magnetic flux through the coil starts to decrease. Consequently, an induced EMF is inverted.
Whenever there is a relative movement between the loop and the magnet, regardless of who moves, an electric current, called induced current, appears in the loop.
See more about magnet at brainly.com/question/13026686
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical nerve cell, 9.0 pC of charge flows in a time of 0.50 ms. Part A What is the average current
Answer:
1.8 x 10^-8A
Explanation:
Using the equation for current:
I= Q/t
current = (9X10^-12)/ (0.50X 10^-3)
= 1.8^-8A
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to (a) (1/3), (b) (1/10)
Answer:
35.3°
18.4°
Explanation:
a.
The first polariser polarises the unpolarised light reducing its intensity from I0 to I0/2. We have to reduce the intensity from I0/2 to I0/3.
Using to Law of Malus, I=I0cos²θ
cos²θ=I/I0=(I0/3)/I0/2 ,
cosθ=√2/3−−√=0.6667−−−−−√=0.8165
θ=cos−1(0.8165)=35.3∘
B.
Cos²θ=I/Io =Io/10/Io9
Cosθ= √9/10= 0.9487
= cos−10.9487
=18.4°
(a) The angle of polaroid such that intensity reduces by 1/3 is 35.26°
(b) The angle of polaroid such that intensity reduces by 1/10 is 63.43°
Angle of polarisation:According to the Malus Law: The intensity of light when passing through a polarizer is given by:
I = I₀cos²θ
where θ is the angle of the polarizer axis with the direction of polarization of the light
I₀ is the initial intensity
When an unpolarised light passes through a polarizer, θ varies from 0 to 2π, so the intensity after passing the first polarizer is :
I = I₀<cos²θ> { average of cos²θ, for 0<θ<2π}
I = I₀/2
Now, this emerging light passes through a second polarizer such that:
(a) the intensity is I' = I₀/3
From Malus Law:
I' = Icos²θ
I₀/3 = (I₀/2)cos²θ
cos²θ = 2/3
θ = 35.26°
(b) the intensity is I' = I₀/10
From Malus Law:
I' = Icos²θ
I₀/10 = (I₀/2)cos²θ
cos²θ = 1/5
θ = 63.43°
Learn more about Malus Law:
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A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________.
A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct
Answer:
4.25m/sE. None of the option is correctExplanation:
Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.
Mathematically.
mu + MU = (m+M)v
m and M are the masses of the bullet and the block respectively
u and U are their respective velocities
v is their common velocity
from the question, the following parameters are given;
m = 20g = 0.02kg
u = 960m/s
M = 4.5kg
U =0m/s (block is at rest)
Substituting this values into the formula above to get v;
0.02(960)+4.5(0) = (0.02+4.5)v
19.2+0 = 4.52v
4.52v = 19.2
Dividing both sides by 4.52
4.52v/4.52 = 19.2/4.52
v = 4.25m/s
Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
1 A diffraction grating has a spacing of 1.6 10-m.
A beam of light is incident normally on the
grating. The first order maximum makes an angle
of 20° with the undeviated beam.what is wavelength of the incident light
Two parallel wires run in a north-south direction. The eastern wire carries 15.0 A northward while the western wire carries 6.0 A northward. If the wires are separated by 30 cm, what is the magnetic field magnitude and direction at a point between the wires at a distance of 10 cm from the western wire?
Answer:
The magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
Explanation:
Given;
current in the eastern wire, [tex]I_e[/tex] = 15 A
current in the western wire, [tex]I_w[/tex] = 6 A
distance between the wires, d = 30 cm = 0.3 m
The magnetic field at a distance R from a line current I, is given as;
[tex]B = \frac{\mu_o I }{2 \pi R}[/tex]
The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.
The total field = magnetic field (east) + magnetic field (west);
[tex]B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})[/tex]
where;
[tex]R_w[/tex] is the distance of the field from west = 10cm = 0.1 m
[tex]R_e[/tex] is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m
The total magnetic field is;
[tex]B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B = 2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T[/tex]
Since total magnetic field is positive, the direction of the field is upwards (positive y direction)
Therefore, the magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3, 2, 1, 4 D. 3, 1, 2, 4
Given that,
Rank in order from largest to smallest the magnitude of the electric field at block dot.
Electric field :
Electric field is proportional to the charge divided by square of distance.
In mathematically,
[tex]E\propto\dfrac{q}{r^2}[/tex]
Where, q = charge
r = distance
If the charge is greater then electric field will be greater.
If the distance is greater then electric field will be smaller.
We need to find the electric field at black dot
According to figure,
(I). The electric field at black dot due to positive charge point q to left. the distance is r.
The electric field will be
[tex]E=\dfrac{kq}{r^2}[/tex]
The electric field will be largest.
(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{2r^2}[/tex]
The electric field will be smallest.
(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(r)^2}[/tex]
The electric field will be very largest.
(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{kq}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{4r^2}[/tex]
The electric field will be very smallest.
So, The electric field from largest to smallest will be
[tex]E_{3}>E_{1}>E_{2}>E_{4}[/tex]
Hence, The ranking will be 3, 1, 2, 4.
(D) is correct option.
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
In what directions is it possible that the wave is traveling?
A. The-z direction.
B. The ty direction
C. The +x direction.
D. The -y direction
E. The -x direction.
F. The +z direction.
Answer:
The wave will be travelling in the y-axis
Explanation:
An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.
define the term DNA
Answer:
It is the carrier of genetic information.
A certain shade of blue has a frequency of 7.06×1014 Hz. What is the energy E of exactly one photon of this light? Planck's constant h=6.626×10−34 J⋅s.
Answer:
Energy, [tex]E=4.67\times 10^{-19}\ J[/tex]
Explanation:
It is given that,
Frequency of blue shade is, [tex]f=7.06\times 10^{14}\ Hz[/tex]
We need to find the energy of exactly one photon of this light. The formula that is used to find the energy of photon is given by :
[tex]E=nhf[/tex]
Here, n is number of photon, n = 1
h is Planck's constant
So,
[tex]E=1\times 6.626\times 10^{-34}\times 7.06\times 10^{14}\\\\E=4.67\times 10^{-19}\ J[/tex]
So, the energy of this light is [tex]4.67\times 10^{-19}\ J[/tex].
The speed of sound in air is 340 m/s, and the density of air is 1.2 kg/m3. If the displacement amplitude of a 330-Hz sound wave is 14 µm, what is its pressure-variation amplitude?
I bel.ieve the answer is 279Ghz
The required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.
Given data:
The speed of sound in air is, v = 340 m/s.
The density of air is, [tex]\rho = 1.2 \;\rm kg/m^{3}[/tex].
The frequency of sound wave is, f = 330 Hz.
The displacement amplitude of sound wave is, [tex]A = 14 \;\rm \mu m= 14 \times 10^{-6} \;\rm m[/tex].
The standard expression for the pressure variation amplitude for the sound wave propagating in air medium is,
[tex]\Delta P= B \times A \times K[/tex]
Here,
B is the Bulk Modulus and its value is, [tex]B = \rho \times v^{2}[/tex].
K is the wave form constant and its value is, [tex]K = \dfrac{2 \pi f}{v}[/tex].
Solving as,
[tex]\Delta P= (\rho \times v^{2}) \times A \times \dfrac{2 \pi f}{v}\\\\\Delta P= (\rho \times v) \times A \times (2 \pi f)\\\\\Delta P= (1.2 \times 340) \times (14 \times 10^{-6}) \times (2 \pi \times 330)\\\\\Delta P= 11.84 \;\rm Pa[/tex]
Thus, we can conclude that the required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.
Learn more about the pressure-variation amplitude here:
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The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 37.8 ~\text{mT}37.8 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 40.640.6 km/s to be undeflected. Give your answer in units of kV/m.
Answer:
50k/h is the answer to iy
To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Answer:
Vi = 2 m/s
Explanation:
First we find the force applied to the car by wall to stop it. We use Hooke's Law:
F = kx
where,
F = Force = ?
k = spring constant = 4 x 10⁶ N/m
x = compression = 3.18 cm = 0.0318 m
Therefore,
F = (4 x 10⁶ N/m)(0.0318 m)
F = 127200 N
but, from Newton's Second Law:
F = ma
a = F/m
where,
m = mass of car = 2010 kg
a = deceleration = ?
Therefore,
a = 127200 N/2010 kg
a = 63.28 m/s²
a = - 63.28 m/s²
negative sign due to deceleration.
Now, we use 3rd equation of motion:
2as = Vf² - Vi²
where,
s = distance traveled = 3.18 cm = 0.0318 m
Vf = Final Speed = 0 m/s
Vi = Initial Speed = ?
Therefore,
2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²
Vi = √4.02 m²/s²
Vi = 2 m/s
Match the followings.
a. The current is induced when there is ________ magnetic flux through a closed loop of wire.
b. If the magnetic flux was constant then there __________ induced current regardless of the magnetic flux value.
c. If the magnetic flux was not constant then there ________ induced current regardless of the magnetic flux value.
1. a constant
2. a strictly decreasing
3. either decreasing or Increasing
4. will be
5. will be no
6. a strictly Increasing
Answer:
a. The current is induced when there is ____3. EITHER DECREASING OR INCREASING___ magnetic flux through a closed loop of wire.
b. If the magnetic flux was constant then there _______WILL BE NO___ induced current regardless of the magnetic flux value.
c. If the magnetic flux was not constant then there __WILL BE______ induced current regardless of the magnetic flux value.
Explanation:
THIS IS BECAUSE IF FARADAY'S LAW OF ELECTROMAGNETIC INDUCTION WHICH STATES THAT WHENEVER THERE IS A CHANGE IN MAGNETIC LINES OF FORCE LINKED WITH A CLOSED CIRCUIT AN EMF IS INDUCED
A circular loop in the plane of a paper lies in a 0.75 T magnetic field pointing into the paper. The loop's diameter changes from 18.0 cm to 6.8 cm in 0.46 s.
A) Determine the direction of the induced current and justify your answer.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
Explanation:
A.the direction of induced current will be clockwise
B: Changing 18cm and 6.8cm into 0.18m and 0.68
2.5
Divide them both by 2 to find the radius . Now we have 0.09 and .034m.
Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb
(π*0.034^2)(.75 T)cos0 and the 0.00272wb
ow use ε=-N(ΔΦ/Δt)
For ΔΦ, 0.091-0.0027=0.0883
C.
To find the current, use I=ε/R
0.0883/2.5= 0.035A
To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface
Answer:
Explanation:
jjjjjjjjjjjjjjjj
An automobile of mass 2500 kg moving at 49.4 m/s is braked suddenly with a constant braking force of 8,868 N. How far does the car travel before stopping
Answer:
344.68 m
Explanation:
The computation of the far does the car travel before stopping is
Data provided in the question
Force = F = 8,868 N
mass = m = 2,500 kg
So,
accleration = a is
[tex]= \frac{-F}{m}\\\\\= \frac{8868}{2500}[/tex]
a = -3.54 m/s^2
The initial speed = u = 49.4 m / s
final speed = v = 0
Based on the above information
Now applying the following formula
v^ 2- u^ 2= 2aS
Therefore
[tex]S = \frac{v^ 2- u^ 2}{2a}\\\\\ = \frac{0- 49.4^ 2}{2\times -3.54}[/tex]
= 344.68 m
A conducting sphere with radius R is charged until the magnitude of the electric field just outside its surface is E. The electric potential of the sphere, relative to the potential for away, is: Group of answer choices 0 E/R E/R2 ER ER2
Answer:
he correct answer is V = ER
Explanation:
In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related
ΔV = ∫ E.ds
where E is the elective field and normal displacement vector.
Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.
ΔV = ∫ E ds
ΔV = E s
since s is in the direction of the radii its value on the surface of the spheres s = R
ΔV = E R
checking the correct answer is V = ER
A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the same elevation 20 m away. Determine the impulse of the club C on the ball.
Answer:
Explanation:
Range of projectile R = 20 m
formula of range
R = u² sin2θ / g
u is initial velocity , θ is angle of projectile
putting the values
20 = u² sin2x 40 / 9.8
u² = 199
u = 14.10 m /s
At the initial point
vertical component of u
= u sin40 = 14.1 x sin 40
= 9.06 m/s
Horizontal component
= u cos 30
At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .
Horizontal component of velocity
u cos 30
Vertical component
= - u sin 30
= - 9.06 m /s
So its horizontal component remains unchanged .
change in vertical component = 9.06 - ( - 9.06 )
= 18.12 m /s
change in momentum
mass x change in velocity
= .050 x 18.12
= .906 N.s
Impulse = change in momentum
= .906 N.s .
The wave number (k) and the angular frequency of a wave are 6.2rad/m and 12rad/s respectively. Which of the followings could be the equation of the wave?
a. 8sin(12x-6.2t)
b. 8sin(6.2x-12t)
c. 12 sin (6.2x-8t)
d. all of the above
Answer:
B. [tex]8sin(6.2x-12t)[/tex]
Explanation:
The general equation of a wave is expressed as [tex]y = Asin(kx-\omega t)[/tex]
A is the amplitude of the wave
k is the wave number and it is expressed as [tex]k =\frac{2\pi}{\lambda}[/tex]
[tex]\omega[/tex] is the angular frequency expressed as [tex]\omega = 2\pi f[/tex]
[tex]\lambda[/tex] is the wavelength and f is the angular frequency
Given k = 6.2rad/m and [tex]\omega = 12rad/s[/tex]
On substituting this value into the general wave equation;
[tex]y = Asin(kx-\omega t)\\y = Asin(6.2x-12t)[/tex]
From the expression gotten, the only equation that could be the equation of the wave is [tex]y = 8sin(6.2x-12t)[/tex]
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.
Answer:
0.29D
Explanation:
Given that
F = G M m / r2
F = GM(6m) / (D-r)2
G Mm/r2 = GM(6m) / (D-r)2
1/r2 = 6 / (D-r)2
r = D / (Ö6 + 1)
r = 0.29 D
See diagram in attached file
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine