Answer:
E
A
C
B
D
Explanation:
A compact fluorescent bulb is 17.0% efficient. How much energy input would be required for the bulb to produce 252 J of light energy?
What time of energy is the waste output and how much would be created?
Help
Answer:
1482.35J
Explanation:
Efficency=(output energy/input energy) x100
17.0 =(252/input energy) x100
input energy =100x252/17=1482.35J
wasted energy=1482.35-252=1230.35J
What would the current be for a circuit that has a voltage of 0.8 V and a resistance of 0.01 Q?
0 1 = 0.01 A
0 1 = 0.8 A
0 1 = 80 A
O I = 0.08 A
Answer:
80 A
Explanation:
Hi there!
Ohm's law states that [tex]V=IR[/tex] where V is the voltage, I is the current and R is the resistance.
Plug the given information into Ohm's law (V=0.8, R=0.01) and solve for I
[tex]V=IR\\0.8=I(0.01)[/tex]
Divide both sides by 0.01 to isolate I
[tex]0.8=I(0.01)\\\frac{0.8}{0.01}= \frac{I(0.01)}{0.01} \\80=I[/tex]
Therefore, the current for this circuit would be 80 A.
I hope this helps!
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
a - is the amount of matter in this object
Which of the following best describes Earth's crust, according to the theory of plate tectonics?
Answer:
the Earth's crust is broken into about 12 plates that float on hotter, softer rocks in the underlying mantle
Explanation:
A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requires 200W of mechanical power. If we assume an efficiency similar to humans (say, 25%), a reasonableassumption, then the metabolic power of the swan is significantly higher thanthis. The swan does not stop to eat during a long day of flying; it get theenergy it needs from fat stores. Assuming an efficiency similar to humans, after12 hours of flight.
Required:
a. How far has the swan traveled?
b. How much metabolic energy has it used?
c. What fraction of its body mass does it lose?
Answer:
Part A:
Distance=864000 m=864 km
Part B:
Energy Used=ΔE=8638000 Joules
Part C:
[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]
Explanation:
Given Data:
v=20m/s
Time =t=12 hours
In Secs:
Time=12*60*60=43200 secs
Solution:
Part A:
Distance = Speed**Time
Distance=v*t
Distance= 20*43200
Distance=864000 m=864 km
Part B:
Energy Used=ΔE= Energy Required-Kinetic Energy of swans
Energy Required to move= Power Required*time
Energy Required to move=200*43200=8640000 Joules
Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]
[tex]K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules[/tex]
Energy Used=ΔE=8640000 -2000
Energy Used=ΔE=8638000 Joules
Part C:
Fraction of Mass used=Δm/m
For This first calculate fraction of energy used:
Fraction of energy=ΔE/Energy required to move
ΔE is calculated in part B
Fraction of energy=8638000/8640000
Fraction of energy=0.99977
Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]
Now, the relation between energies ratio and masses is:
[tex]\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2[/tex]
[tex]\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977[/tex]
[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]
One way to probe the nucleus is to bombard a sample with high-energy electrons. To learn about the nuclear structures in a sample, the de Broglie wavelengths of these electrons would need to be a little smaller than a nuclear radius. Estimate the energy of such electrons. Give your answer in electron-volts. (Assume that the wavelength used is about 9.0 fm.) eV
Answer:
E = 1.38 x 10⁸ eV = 138 MeV
Explanation:
The energy associated with the given wavelength can be found from the following formula:
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = Energy of electron = ?
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = Speed of Light = 3 x 10⁸ m/s
λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m
Therefore,
[tex]E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})[/tex]
E = 1.38 x 10⁸ eV = 138 MeV
Which of the following should a warm up NOT include?
Select one:
A. A few minutes of cardiovascular exercise
B. A light version of the exercise you will be performing
C. Intense stretching
D. All of the above should be included in a warm up
Which of these properties of light is a constant?
speed in a vacuum
amplitude
wavelength
frequency
Answer:
speed in vacuum
Explanation:
lets say we are in an empty universe and you are moving 10% the speed of light you wont slow down or speed up.
Choose the correct statement regarding the sign conventions for lenses.
a. The focal length f is positive for diverging lenses.
b. Virtual images appear on same side of the lens as the object and have a negative value for the image distance.
c. Real images appear on the opposite side of the lens from the object and have a negative value for the image distance.
d. The focal length f is negative for converging lenses.
e. Virtual images appear on same side of the lens as the object and have a positive value for the image distance.
Answer:
a) false
b) True
c) True
d) False
e) False
Explanation:
a) False
For a diverging lens, the focal length is negative while it is positive for a converging lens
b) True
Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged
c) True
d) False
For a diverging lens, the focal length is negative while it is positive for a converging lens
e) False
Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyond the edge hanging freely in air. how far beyond the edge of the wall can a 100kg woman walk before the plank began to rotate about the edge of the wall
Answer:
x₂ = 1.33 m
Explanation:
For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall
Στ = 0
W₁ x₁ - W₂ x₂ = 0
where W₁ is the weight of the woman, W₂ the weight of the table.
Let's find the distances.
Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.
x₁ = 2.5 -1.5 = 1 m
The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative
x₂ = [tex]\frac{M_1g }{m_2 g} \ x_1[/tex]
let's calculate
x₂ = [tex]\frac{100}{75} \ 1[/tex]
x₂ = 1.33 m
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 1.61THz. Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.
Answer:
λ = 1.86 x 10⁻⁴ m = 186 μm
Explanation:
The relationship between the wavelength and the frequency of a wave is given by the following equation:
[tex]c = f\lambda\\\\\lambda = \frac{c}{f}[/tex]
where,
λ = wavelength of infrared radiation = ?
c = speed of infrared radiation = speed of light = 3 x 10⁸ m/s
f = frequency of infrared radiation = 1.61 THz = 1.61 x 10¹² Hz
Therefore,
[tex]\lambda = \frac{3\ x\ 10^8\ m/s}{1.61\ x\ 10^{12}\ Hz}[/tex]
λ = 1.86 x 10⁻⁴ m = 186 μm
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?
Answer:
λ = 5.734 x 10⁻⁷ m = 573.4 nm
Explanation:
The formula of the Young's Double Slit experiment is given as follows:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x d}{L}[/tex]
where,
λ = wavelength = ?
L = distance between screen and slits = 8.61 m
d = slit spacing = 1.09 mm = 0.00109 m
Δx = distance between consecutive bright fringes = [tex]\frac{4.53\ cm}{10}[/tex] = 0.00453 m
Therefore,
[tex]\lambda = \frac{(0.00453\ m)(0.00109\ m)}{8.61\ m}[/tex]
λ = 5.734 x 10⁻⁷ m = 573.4 nm
Two vectors have magnitudes 3 and 4 . how are the directions of the two vectors related if: a/the sum has magnitude 7.0
A child holds one end of a 33.0-meter long rope in her hand and moves it up-and-down to produce a sinusoidal wave by moving her hand from 8.00 cm above her shoulder to 8.00 cm below her shoulder at a frequency of 2.00 Hz and a wavelength of 75.0 cm. If the child doubles the amplitude of her hand's motion on that same rope with the same tension in the rope, then what will the wavelength of the wave now be
Answer:
Wavelength=75 cm.
The wavelength well remain unchanged which is 75 cm.
Explanation:
The formula which will help us to answer the question is:
V=f*λ
Where:
V is the velocity
f is the frequency of wave
λ is the wave length
Now:
λ=V/f Eq (1)
The equation show's that wavelength is independent of the amplitude but it depends on the frequency and the velocity with which wave is moving.
The wavelength well remain unchanged which is 75 cm.
In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- vestigated occurs when the external field Bext has magnitude 0.78 T. Assume that Bint and Bext are in the same direction and take the pro- ton magnetic moment component u, to be 1.41 X 10-26 J/T. What is the magnitude of Bint?
Answer:
[tex]B_{int}=-0.015T[/tex]
Explanation:
From the question we are told that:
RF source oscillation speed [tex]\sigma= 34 MHz[/tex]
The external field [tex]Bext =0.78 T[/tex].
Pro- ton magnetic moment component [tex]\mu=1.41 X 10-26 J/T[/tex]
Generally the equation for magnitude of [tex]B_{int}[/tex] is mathematically given by
[tex]B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}[/tex]
[tex]B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}[/tex]
[tex]B_{int}=0.78-0.7957[/tex]
[tex]B_{int}=-0.015T[/tex]
An experiment is performed on an unknown material and produces the given heat curve. The temperature of the material is shown as a function of heat added. Other experiments determine that the material has a temperature of fusion of fusion=235 °C and a temperature of vaporization of vapor=471 °C.
If the sample of material has a mass of =9.80 g, calculate the specific heat when this material is a solid, s, and when it is liquid, l.
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
In this case, we need to determine that specific heat for solid and liquid states of matter. By Heat Physics, we understand that specific heat is contained in the slopes of the two sensible phases in the following form:
[tex]\frac{\Delta T}{\Delta Q} = \frac{1}{m\cdot c}[/tex] (1)
Where:
[tex]\Delta T[/tex] - Temperature change, in degrees Celsius.[tex]\Delta Q[/tex] - Heat received, in joules.[tex]m[/tex] - Mass of the sample, in grams.[tex]c[/tex] - Specific heat of the sample, in joules per kilogram-degrees Celsius.Solid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{1} = 40\,^{\circ}C[/tex], [tex]T_{2} = 235\,^{\circ}C[/tex], [tex]Q_{1} = 183\,J[/tex] and [tex]Q_{2} = 819\,J[/tex], then the specific heat of the solid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{819\,J-183\,J}{(9.80\,g)\cdot (235\,^{\circ}C - 40\,^{\circ}C)}[/tex]
[tex]c = 0.333\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
Liquid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{3} = 230\,^{\circ}C[/tex], [tex]T_{4} = 471\,^{\circ}C[/tex], [tex]Q_{3} = 1470\,J[/tex] and [tex]Q_{4} = 2870\,J[/tex], then the specific heat of the liquid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{2870\,J - 1470\,J}{(9.80\,g)\cdot (471\,^{\circ}C - 230\,^{\circ}C)}[/tex]
[tex]c = 0.593\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
We kindly invite you to see this question related to specific heat: https://brainly.com/question/11194034
A sound wave has a wavelength of 2M and a frequency of 100 Hz. The speed of the wave is
Answer:
200 meters per second
Explanation:
wave speed = wavelength x frequency
2 meters x 100 hertz = 200 meters per second
What is the frequency of a wave traveling at 300,000,000 m/s with a wavelength of .0025 m/cycle? Please help me !!
Answer:
its 00.0035474
Explanation:
..... .. . . .. . . . . .. .
Object X of mass 4 kg travels with a speed of 3 ms toward object Y of mass 2 kg that is initially at rest. Object X then collides with and sticks to object Y . After the collision, object X and object Y remain stuck together. How much mechanical energy is converted into nonmechanical energy during the collision?
The nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]
What is the conservation of momentum?The conservation of the momentum is defined as when two bodies collide with each other then the total energy of the masses will remain constant.
It is given in the question that
Mass of the body X is [tex]M_X=4\ kg[/tex]
The velocity of the body X is =[tex]V_x=3 \ \frac{m}{s}[/tex]
Mass of the body Y is [tex]M_Y= 2\ kg[/tex]
The velocity of body Y is [tex]V_y= 0[/tex]
Now to find out the energy converted after the collision we will first find the final velocity of both the bodies.
Now from the conservation of the momentum
[tex]M_X V_X +M_YV_Y=(M_X+M_Y)V_F[/tex]
[tex](4\times 3)+(2\times0)=(4+2)\times V_F[/tex]
[tex]V_F= \dfrac{12}{6}[/tex]
[tex]V_F = 2 \frac{m}{s}[/tex]
Now to find the Change in the energy of the body
[tex]\Delta E=E_i-E_f[/tex]
[tex]\Delta E=\dfrac{1}{2} M_XV_X^2+\dfrac{1}{2} M_YV_Y^2-\dfrac{1}{2} (M_X+M_Y)V_F^2[/tex]
[tex]\Delta E=\dfrac{1}{2} (4)(3^2)+\dfrac{1}{2} (2)(0^2)-\dfrac{1}{2} ( 4+2) 2^2[/tex]
[tex]\Delta E= 6\ J[/tex]
Thus the nonmechanical energy or the loss of energy after the collision will be equal to [tex]E=6\ J[/tex]
To know more about the collision follow
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Conservation equations are powerful equations, because of how versatile and flexible they are to different physical situations. Conservation of mass is a simple one that can even be applied to fluids with care. The conservation of momentum also applies to any closed system, such as collisions. We've examined the nuances of energy conservation in our labs with total mechanical energy. Does kinetic energy remain conserved in all collisions too?
a) true
b) false
Answer:
b) false
Explanation:
Since in the given situation it is mentioned that the mass conversation would be simple and same applied to the fluids with care also the conservation of momentum would be applied to any type of closed system like collisions
But as we know that
In the inelastic collisions, the total kinetic energy would not be remain conserved
So the given statement is false
PLS HELP.
A rope breaks when the speed of a 0.309 kg mass moving in a circle of radius of 0.429 m reaches 12.9 m/s
how much tension is in the strings when it breaks?
unti=n
120 works for acellus
Answer: 119.9
Explanation:
F = (mv^2)/r
Here we know m (0.309kg), v (12.9m/s) and r (0.429m)
So F = (0.309*12.9^2)/0.429 = 119.861748
One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the 12 C and 13 C isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 km/s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the 12 C. The measured masses of these isotopes are 1.99×10−26kg(12C) and 2.16×10−26kg(13C).
(a) What strength of magnetic field is required?
(b) What is the diameter of the 13 C semicircle?
(c) What is the separation of the 12 C and 13 C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?
Answer:
[tex]0.0084575\ \text{T}[/tex]
[tex]0.272\ \text{m}[/tex]
2.2 cm easily observable
Explanation:
[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]
[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]
[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]
B = Magnetic field
v = Velocity of atom = 8.5 km/s
[tex]r_2[/tex] = Radius of 13 C
The force balance of the system is
[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]
The required magnetic field is [tex]0.0084575\ \text{T}[/tex]
Radius is given by
[tex]r=\dfrac{mv}{qB}[/tex]
[tex]r\propto m[/tex]
So
[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]
The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]
Separation is given by
[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]
The distance of separation is 2.2 cm which is easily observable.
Jack has a weight of 300 N and sits 2.0 m from the pivot of see - saw. Jill has a weight of 450 N and sits 1.5 m from pivot. Who will move down?
Answer:
Jill will move down first
Explanation:
A football is thrown down off of a building with a force of 70
Newtons. Gravity is pulling down the football with a force of
19 Newtons. What is the net force on the football?
Answer:
89 N downward.
Explanation:
Both forces are working in the same direction -- down.
Therefore you can add them together.
F = F1 + F2
F is the total force
F1 is the gravitational force
F2 is the force the person provides
F1 = 19
F2 = 70
F = ?
F = 19 + 70
F = 89 Newtons downward.
Answer:
wE dONt taLk aBt brunooooOoooOoOo ;')
Explanation:
Which of the following types of electromagnetic radiation has waves with the highest frequency ? A. Infrared radiation B. Visible light c. Microwaves D. X-rays
Which statement is FALSE? *
explosives that deflagrate contain explosive materials that react (burn) slower than the speed of sound
an example of a deflagration explosive is C4
some deflagration explosives have shock waves that move faster than the speed of sound
explosives that deflagrate are also called low explosives
Answer:
Some deflagration explosives have shock waves faster than the speed of sound.
Explanation:
A example of a deflagration explosives.
Answer:
an example of a deflagration explosive is C4, is false.
Explanation:
Explosive materials may be categorized by the speed at which they expand. Materials that detonate (the front of the chemical reaction moves faster through the material than the speed of sound) are said to be "high explosives" and materials that deflagrate are said to be "low explosives".
C4 meaning:
C-4 or Composition C-4 is a common variety of the plastic explosive family known as Composition C, which uses RDX as its explosive agent. C-4 is composed of explosives, plastic binder, plasticizer to make it malleable, and usually a marker or odorizing taggant chemical.
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 805 N and stands 1.00 m from the left end. Two meters from the left end is the 500 N washing equipment. Joe is 0.500 m from the right end and weighs 820 N. Given that the scaffold is in rotational and translational equilibrium, what are the forces on each cable
Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ (805 N)( (3-1) m) + ( 395 N )( [tex]\frac{3}{2} m[/tex]) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ 1610 + 592.5 + 500 + 410 ]
[tex]T_{left[/tex] = [tex]\frac{1}{3m}[/tex][ 3112.5 ]
[tex]T_{left[/tex] = 1037.5 N
[tex]T_{left[/tex] = 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
[tex]T_{Right[/tex] = ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
[tex]T_{Right[/tex] = 2520 N - 1037.5 N
[tex]T_{Right[/tex] = 1482.5 N
[tex]T_{Right[/tex] = 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Calculate the decay constant of the radioisotope.
Answer: [tex]0.0353\ s^{-1}[/tex]
Explanation:
Given
Radioactive material is found to decrease 40% of its original value in [tex]2.59\times 10\ s[/tex]
Sample at any time is given by
[tex]N=N_oe^{-\lambda t}[/tex]
where, [tex]\lambda=\text{decay constant}[/tex]
Put values
[tex]\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10[/tex]
Taking natural logarithm both side
[tex]\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}[/tex]
What makes astronomers think that impact rates for the Moon must have been higher earlier than 3.8 billion years ago?
Answer: See explanation
Explanation:
The reason why astronomers think that the rates of impact for the Moon must have been higher earlier than 3.8 billion years ago is because on the older highlands, there are ten times more craters than on the younger maria.
It is believed that the impact rate was higher earlier and thus can be seen when the numbers of the craters that can be seen on the lunar highlands is being compared to that on the maria. It should be noted that there are about 10 times more craters that can be found on the highlands than those on the maria.
If there was a constant rate of impact throughout the history of the Moon, then the highlands be about 10 times older and therefore will have been formed about 38 billion years ago.