Solution :
Given data:
Area of the solar sail, A = 16 [tex]$m^2$[/tex]
Mass (array + sail), m = [tex]$3 \times 10^{-3}$[/tex] kg
Power, P = 100 GW
= [tex]$10^{11}$[/tex] W
Time, t = 10 min
= 10 x 60 s
Distance, D = [tex]$4.3 \times 4 \times 10^{16}$[/tex] m
Kinetic energy, [tex]$KE=\frac{1}{2}mv^2=P\times t$[/tex]
[tex]$v=\sqrt{\frac{2Pt}{m}}$[/tex]
[tex]$v=\sqrt{\frac{2\times 10^{11}\times 600}{3 \times 10^{-3}}}$[/tex]
[tex]$=2 \times 10^8$[/tex] m/s
So, the acceleration is [tex]$a=\frac{v}{t}$[/tex]
[tex]$a=\frac{2 \times 10^8}{6 \times 10^2} \ m/s^2$[/tex]
[tex]$=333333.33 \ m/s^2$[/tex]
Therefore, force,
[tex]$F = ma$[/tex]
[tex]$=3 \times 10^{-3}\times 333333.33$[/tex]
= 1000 N
Pressure, [tex]$P=\frac{F}{A}$[/tex]
[tex]$=\frac{1000}{16}$[/tex]
[tex]$=62.5 \ N/m^2$[/tex]
Therefore, time taken is t
Now the distance is
[tex]$d_1=\frac{1}{2}at_1^2$[/tex]
[tex]$d_1=0.5 \times 333333.33 \times (600)^2$[/tex]
[tex]$=6 \times 10^{10} \ m$[/tex]
Now, the distance, [tex]$d_2 = D-d_1=v.t_2$[/tex]
Now, [tex]$t_2=\frac{(17.2 \times 10^{16})-(6 \times 10^{10})}{2 \times 10^8}$[/tex]
= 859999700 s
Therefore, total time is
[tex]$T=t_1+t_2$[/tex]
= 86000300 s
= 27.27 years
Imagine you could travel to the moon where the acceleration due to gravity is 1.6 m/s^2. What would be the period of
a pendulum that is 1.0 m?
Show your work.
Answer:
4.9612 s
Explanation:
Applying,
T = 2π√(L/g)............... Equation 1
Where T = period of the pendulum, L = Lenght of the pendulum, g = acceleration due to gravity of the moon, π = pie.
From the question,
Given: L = 1 m, g = 1.6 m/s²
Constant: π = 3.14
Substitute these values into equation 1
T = 2×3.14×√(1/1.6)
T = 6.28√(0.625)
T = 6.28×0.79
T = 4.9612 s
A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the frequency of the echo is 900 Hz higher than the frequency the bat is emitting. The speed of sound in air is 340 m/s at emits a sound at a frequency of 30.0 kHz
(a) What is the speed of the bat?
(b) What is the wavelength of the sound that the bat hears15?
Answer:
a) the speed of the bat is 5.02 m/s
b) the wavelength of the sound that the bat hears is 0.011 m
Explanation:
Given the data in the question;
Frequency of sound emitted by a bat f = 30.0 kHz = 30000 Hz
detected frequency by the bat δf = 900 Hz
speed of sound in air c = 340 m/s
Let speed of sound and speed of bat be c and [tex]v_s[/tex] respectively;
Now, frequency of the sound that is coming from the bat towards the wall due to DROPPLER EFFET will be;
f₁ = ( c / ( c - [tex]v_s[/tex] ) )f ----- let this be equ 1
Also, frequency does not change after deflection. The bat becomes an observer as the dropper is shifted because the reflected sound wave is coming towards it;
Hence, Doppler shifted frequency will be;
f₂ = ( (c + [tex]v_s[/tex] ) / c )f₁
from equ 1, f₁ = ( c / ( c - [tex]v_s[/tex] ) )f, so we substitute
f₂ = ( (c + [tex]v_s[/tex] ) / c ) × ( c / ( c - [tex]v_s[/tex] ) )f
f₂ = ( (c + [tex]v_s[/tex] ) / ( c - [tex]v_s[/tex] ) )f
∴ beat frequency will be;
δf = f₂ - f = ( (c + [tex]v_s[/tex] ) / ( c - [tex]v_s[/tex] ) )f - f
δf = ( 2[tex]v_s[/tex] / c - [tex]v_s[/tex] )f
δf = ( 2[tex]v_s[/tex] / c - [tex]v_s[/tex] )f
2f/δf = c - [tex]v_s[/tex] / [tex]v_s[/tex]
2f/δf = c/[tex]v_s[/tex] - [tex]v_s[/tex] / [tex]v_s[/tex]
2f/δf = c/[tex]v_s[/tex] - 1
c/[tex]v_s[/tex] = 2f/δf + 1
[tex]v_s[/tex] = c / (2f/δf + 1)
now, we substitute in our values;
[tex]v_s[/tex] = 340 / ((2×30000 / 900 ) + 1)
[tex]v_s[/tex] = 340 / (66.6666 + 1)
[tex]v_s[/tex] = 340 / 67.6666
[tex]v_s[/tex] = 5.02 m/s
Therefore, the speed of the bat is 5.02 m/s
b) the wavelength of the sound that the bat hears
frequency of reflected wave is;
f₂ = f + δf = 30000 + 900 = 30900 Hz
λ₂ = c / f₂
we substitute
λ₂ = 340 / 30900
λ₂ = 0.011 m
Therefore, the wavelength of the sound that the bat hears is 0.011 m
For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples of 5 units has been 3.4 volts, and the range has been 1.3 volts.
Required:
What would the upper and lower control limits be for the resulting control charts (average and range)?
Answer:
Average :
UCL = 4.15
LCL = 2.65
Range :
UCL = 2.75
LCL = 0
Explanation:
Given :
Sample size, n = 5
Average, X = 3.4
Range, R = 1.3
A2 for n = 5 ; equals 0.577 ( X chart table)
For the average :
Upper Control Limit (UCL) :
X + A2*R
3.4 + 0.577(1.3) = 4.1501
Lower Control Limit (LCL) :
X - A2*R
3.4 - 0.577(1.3) = 2.6499
FOR the range :
Upper Control Limit (UCL) :
UCL = D4*R
D4 for n = 5 ; equals = 2.114
UCL = 2.114*1.3 = 2.7482
Lower Control Limit (LCL) :
LCL = D3*R
D3 for n = 5 ; equals = 0
LCL = 0 * 1.3 = 0
What is the period of a wave, with a frequency of 0.75 Hertz?
Answer:
wavelength 4 cm
Explanation:
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, lowmass sail and the energy and momentum of sunlight for propulsion.
a. Should the sail be absorbing or reflective? Why?
b. The total power output of the sun is 3.9×10^26W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers.
c. Explain why your answer to part (b) is independent of the distance from the sun.
Answer:
a) the reflective surface has twice the energy transfer
b) A = 1.3 10²⁷ km²
c) the energy emitted by the sun is distributed in a sphere that depends on the square of the distance, and the gravitational force depends on the square of the distance
Explanation:
a) The pressure exerted on the candle is related to the variation of the momentum
P = [tex]\frac{1}{c} \ \frac{dp}{dt}[/tex]
in a case of absorption (inelastic shock) all the energy is absorbed therefore the pressure is
P = \frac{1}{c} \ \frac{dp}{dt}
in the case of reflection (elastic shock) an energy is absorbed by absorbing the light and then by action and reaction the same energy is absorbed in the reflected light
P = 2 \frac{1}{c} \ \frac{dp}{dt}
In conclusion, the reflective surface has twice the energy transfer.
b) pressure is defined with force per unit area
P = F / A
F = P A
this force must be greater than the gravitational force of attraction of the sun
Fg = G m Ms / r²
let's look for the case that the two forces are equal
F = Fg
P A_sail = G m Ms = r²
suppose a fully reflective sail
[tex]2 \frac{S}{c} \ A_{sail} = G \frac{m M_s}{r^2}[/tex]
The pointing vector is the power delivered per unit area
S = I = P / A
where A is the area of the sphere where the is distributed by the sun
A = 4π r²
we substitute
[tex]\frac{2P}{c} \ \frac{A_{sail}}{4 \pi r^2} = G \frac{m M_s}{r^2}[/tex]
[tex]\frac{1}{2 \pi \ c }[/tex] A_{sail} = G m M_s
A = G m M_s 2π c
let's calculate
A = 6.67 10⁻¹¹ 10000 2 10³⁰ 2π 3 10⁸
A = 1,257 10³³ m²
let's reduce to km²
A = 1.3 10³³ m² (1km / 10³ m) ²
A = 1.3 10²⁷ km²
c) The size of the candle is independent of the distance to the sun because the energy emitted by the sun is distributed in a sphere that depends on the square of the distance, and the gravitational force depends on the square of the distance, therefore the two dependencies are canceled.
how hot can the desert get
Answer:
134 f
Explanation:
The hottest temperature ever reliably measured in a desert was 134 degrees F, in Death Valley of the Mojave Desert in 1913.
Temperature. During the day, desert temperatures rise to an average of 38°C (a little over 100°F). At night, desert temperatures fall to an average of -3.9°C (about 25°F). At night, desert temperatures fall to an average of -3.9 degrees celsius (about 25 degrees fahrenheit).
Desert surfaces receive a little more than twice the solar radiation received by humid regions and lose almost twice as much heat at night. Many mean annual temperatures range from 20-25 degrees Celsius. The extreme maximum ranges from 43.5-49 degrees Celsius. Minimum temperatures sometimes drop to -18 degrees Celsius.
Smoke detectors fall into two major classes. Ionization detectors, the most common units, contain two parallel electrodes that are typically separated by 3 cm with a 5-V potential difference across them. The air molecules between the electrodes are ionized by collisions with helium nuclei that are produced by a radioactive source. Most units are initially fueled with 60 million nuclei of radioactive americium 241 (half-life 430 years). The now-ionized air molecules drift toward one of the electrodes with an average speed of 0.1 m/s and thus support a small current between the two electrodes. Smoke particles that enter and combine with the ions reduce the current and initiate an alarm.
Photoelectric detectors, by contrast, contain a light-emitting diode that sends a beam of unpolarized light across a small chamber. The light beam usually has a wavelength of 6.0 × 10–7m and has an intensity of 1.0 × 10–3 W. When smoke particles enter the chamber, the light scatters in all directions. A photocell then senses either the increase in the scattered light or the reduced intensity of the light beam and sets off the alarm. The speed of light in air is 3.0 × 108 m/s.
Ionization detectors respond faster to the large smoke particles of flaming fires; photoelectric detectors sense the small particles of smoldering fires more quickly. Modern units have both types of detectors.
When fewer than 3.75 × 106 americium nuclei remain, the ionization smoke detector will not operate due to insufficient ionization. How much time will pass before there are this many nuclei remaining?
a. 1720 years
b. 2150 years
c. 4300 years
d. 6880 years
Answer:
1720 years is the amount of time that will pass.
Option a) 1720 years is the correct answer
Explanation:
Given the data in the question;
Number of nuclei initially N₀ = 60 million = 60,000,000
After time t, Number of nuclei remaining N[tex]_{rem[/tex] = 3.75 × 10⁶
Also given that; half-life of radioactive americium [tex]t_{1/2[/tex] = 430 years.
so;
λ = ln2 / [tex]t_{1/2[/tex]
we substitute
λ = ln2 / 430 years
N[tex]_{rem[/tex] = N₀e^(-λt)
solve for t
t = 1/λ × ln( N₀/N[tex]_{rem[/tex] )
so we substitute
t = 1 / (ln2 / 430 years) × ln( 60,000,000 / ( 3.75 × 10⁶ ) )
t = ( 430 years / ln2 ) × ln( 60,000,000 / ( 3.75 × 10⁶ ) )
t = ( 430 years / 0.693147 ) × ln( 16 )
t = 620.359 years × 2.7725887
t = 1720.0003 ≈ 1720 years
Therefore, 1720 years is the amount of time that will pass.
Option a) 1720 years is the correct answer
Magnesium hydroxide is a common ___?
Answer:
Hello There!!
Explanation:
The answer is=> It is a common component of antacids.
hope this helps,have a great day!!
~Pinky~
A series circuit is set up with an AA battery along with an mystery material and ammeter; however, there’s no current passing through.
A.Insulator
B.Conductor
C.Semiconductor
Answer:
A. Insulator
Explanation:
Since there is no current passing through at all.
1. How would the forces from a header with such a light soccer ball cause a concussion? Draw
Two free body diagrams showing how the amount of peak force on the head would compare
to the amount of peak force on a soccer ball in a header that causes a collision.
Answer:
soccer when the ball hits an unprepared player in the head. He also gave examples of concussions occurring when players accidentally knock their heads into other players while attempting to head the ball, particularly if they are attempting to flick the ball backwards.
Explanation:
Heading in soccer can increase your risk of concussions. Over time, repeated subconcussive injuries can also accumulate and cause brain damage.
A 300-g chunk of ice (of density 0.900 g/cm3) is placed in a water bucket. A 20.0-g rock, with a volume of 2.00 cm3, is placed on top of the ice before water is filled to the top. When the ice melts and the rock drops to the bottom of the buckt, how much water spills out or needs to be added to maintain full level? Ignore the possible dependence of the densities of ice and water on temperature.
Answer:
18 cm³ of water
Explanation:
The correct procedure and explanation is in the picture attached. If you have any doubts, feel free to leave in the comments.
Hope this helps
40 points! Will give brainliest!
Which of the following describes an advantage of AC electricity over DC electricity?
A) AC is found in most low voltage operations.
B) AC is provided as strong, short bursts of electricity.
C) AC can be transported over long distances.
D) AC can be used is small electronic devices.
Describe the formation of galaxies.
Today's cosmologists assume that matter was not uniformly distributed in the universe after the Big Bang. Dense places attract more matter than the surrounding area according to their gravitational forces. Over the course of billions of years, these gas agglomerations eventually led to the formation of the galaxies we see today.
5. A ball weighing 10 kg rolls 200 m down a frictionless incline with a 50 degree angle to the horizontal. If the ball’s initial velocity was 0 m/s, how much does the mechanical energy of the system change by the time the ball reaches its destination? A) It increased by 12%. B) It increases by 58%. C) It decreases by 12%. D) It does not change.
Answer:
D) It does not change
Explanation:
Since there is no friction in the inclined plane. Therefore, there is no loss in the total mechanical energy of the system. So according to the law of conservation of energy we can write:
Total Mechanical Energy at Start = Total Mechanical Energy at End + Frictional Loss
Total Mechanical Energy at Start = Total Mechanical Energy at End + 0
Total Mechanical Energy at Start = Total Mechanical Energy at End
It means there is no change in the total mechanical energy of the system.
Therefore, the correct option is:
D) It does not change
what is the focalength of the combination of two thin renses of power -5D and -2D placeci in contact with each other?
Answer:
f = - 0.143 m = - 14.3 cm
Explanation:
First, we will calculate the power of the combination of lenses:
P = P₁ + P₂
where,
P = Power of Combination = ?
P₁ = Power of first lens = - 5D
P₂ = Power of second lens = - 2D
Therefore,
P = - 5D - 2D
P = - 7D
Now, the focal length can be given as:
[tex]f = \frac{1}{P} \\\\f = \frac{1}{-\ 7\ D} \\\\[/tex]
f = - 0.143 m = - 14.3 cm
Negative focal length indicates that combination will act as diverging lens.
What is gravitational force? -,-
Answer:
The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is called gravitational force.
the official world land speed record is1228.0km/h are on October 15,1997,by Andy green in the jet engine car then express the speed in the meters per second
Answer:
v = 341.11 m/s
Explanation:
Given that,
The official world land speed was recorded as 1228.0km/h on October 15,1997.
We need to find the speed in m/s.
We know that,
1 km = 1000 m
1 h = 3600 s
So,
[tex]v=1228\ km/h\\\\=1228\times \dfrac{1000\ m}{3600\ s}\\\\v=341.11\ m/s[/tex]
So, the required speed is 341.11 m/s.
Explain how newtons first law applies to the image to the left.
Aluminium is produced by the reduction of aluminium oxide
What is meant by the term reduction?
Answer:
Explanation:The word reduction in terms of chemistry means loss of Oxygen,Gain of hydrogen,Gain of electrons and loss of oxidation state.
Please mark me as brainiest.
Aluminium is extracted from aluminium oxide by electrolytic reduction.
What is reduction?The process is termed as a reduction if
there is a loss of oxygen or an electronegative atom.there is a gain of hydrogen or an electropositive atom.loss of electrons.decrease in oxidation number.What is electrolytic reduction?The electrolysis method used to obtain metals from molten chloride or oxides is known as electrolytic reduction.
What is the use of electrolytic reduction?It is used in the extraction of highly electropositive metals.
To learn more about reduction and electrons here,
https://brainly.com/question/14698511
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An object of mass 25kg is falling from the height h=10 m. calculate
a. The total energy of an object at h=10m.
b. Potential energy of the object when it is at h= 4m
c. Kinetic energy of the object when it is at h= 4m
d. What will be the speed of the object when it hits the ground?
Answer:
a=2500J,b=1000K,c=1000J,d=14.142m/s
Explanation:
V²=U²+2gh
V²=0 + 2×10×10=200m/s
a).kinetic energy=(1/2)mv²=(1/2)25×200=2500
potential energy=mgh
p.e=25×10×10=2500J
pe+ke=2500+2500=5KJ
b).mgh=25×10×4=1000J
c). V²=U²+2gh
V²=0+2×10×4
V²=80
kinetic energy=(1/2)mv²
=(1/2)25×80
=1KJ
d). From my first paragraph V²=200
V=√200
V=14.142m/s
What is the speed of the wave above of the frequency is 7.0 hertz
Answer:
Umm
Explanation:
Question 1
2 pts
Explain what causes a solution to be a strong acid.
Answer:
Cuanto más fuerte es el ácido, más rápido se disocia para generar H +start superscript, plus, end superscript. Por ejemplo, el ácido clorhídrico (HCl) se disocia completamente en iones hidrógeno y cloruro cuando se mezcla con agua, por lo que se considera un ácido fuerte.
Q: Winding roads and sharp curves are inherently dangerous
because -
A- You can't see far ahead
B- The speed limit is so slow
C- There may be other vehicles traveling in the
opposite direction
D- You may struggle to stay at the same speed
Answer:
A) you can't see far ahead
D. You may struggle to stay at the same speed.
Winding roads and sharp curves are dangerous because of centrifugal forces while taking curves, whose classic formula is described below:
[tex]F = m\cdot \frac{v^{2}}{r}[/tex] (1)
Where:
[tex]F[/tex] - Centrifugal force, in newtons.[tex]v[/tex] - Speed, in meters per second.[tex]r[/tex] - Radius of curvature, in meters per second.Sharps curves are characterized by low radii of curvature. Given that centrifugal force is inversely proportional to the square radius of curvature and it is opposed to friction force between tires and pavement, if centrifugal force is higher than maximum static friction force, then a car accident could if driver does not decelerate.
Therefore, correct choice is D.
We kindly invite to see this question on centrifugal forces: https://brainly.com/question/545816
In a single-slit experiment, the slit width is 150 times the wavelength of the light.
What is the width (in mm) of the central maximum on a screen 2.6 m behind the slit?
I have tried:
y=[(1+1/2)(lambda)(2.6m)] / (150lambda)
to bring me to
y=[(1.5)(2.6)] \ (150)
giving me a y value in mm of 26
A box has base dimensions of 30 cm x 30 cm and a mass of 3 kg. Calculate to what
depth it sinks when placed in a tank of water (assume the sides are sufficiently high,
so that it floats). Density of water = 1000 kgm-3. Assume g=10 ms-2 for this problem.
Answer:
60
Explanation:
A box has base dimensions of 30 cm x 30 cm and a mass of 3 kg. Calculate to what
depth it sinks when placed in a tank of water (assume the sides are sufficiently high,
so that it floats). Density of water = 1000 kgm-3. Assume g=10 ms-2 for this problem.
what are conductors and insulators
A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light intensity is Io, the average energy density of the waves is u0, and the rms electric and magnetic field values are Eo and Bo, respectively.
Required:
a. What is the light intensity?
b. What is the average energy density of the waves?
c. What is the rms magnetic field value?
Answer:
The appropriate solution is:
(a) [tex]\frac{1}{4}(I_o)[/tex]
(b) [tex]\frac{1}{4} (u_o)[/tex]
(c) [tex]\frac{1}{2}B_o[/tex]
Explanation:
According to the question, the value is:
Power of bulb,
= 60 W
Distance,
= 1.0 mm
Now,
(a)
⇒ [tex]\frac{I}{I_o} =\frac{r_o_2}{r_2}[/tex]
On applying cross-multiplication, we get
⇒ [tex]I=I_o\times \frac{1_2}{2^2}[/tex]
⇒ [tex]=I_o\times \frac{1}{4}[/tex]
⇒ [tex]=\frac{1}{4} (I_o)[/tex]
(b)
As we know,
⇒ [tex]\frac{u}{u_o} =\frac{I}{I_o}[/tex]
By putting the values, we get
⇒ [tex]u=\frac{1}{4}(u_o)[/tex]
(c)
⇒ [tex]\frac{B^2}{B_o^2} =\frac{u}{u_o}[/tex]
[tex]=\frac{I}{I_o}[/tex]
⇒ [tex]B=B_o\times \sqrt{\frac{1}{4} }[/tex]
⇒ [tex]=\frac{1}{2}(B_o)[/tex]
When 1.5 kg of mass turns into energy, how much energy is released? Find the equation, substitution, and number with units.
Explanation:
[tex]from \: einstein \: equation : \\ E = m {c}^{2} \\ E = 1.5 \times {(3 \times {10}^{8} )}^{2} \\ E = 1.35 \times {10}^{17} \: joules[/tex]
If a body of mass 2 kg is moving with a velocity of 30 m/s, then
on doubling its velocity the momentum becomes
a 30 kgm/s
b 90 kgm/s
C 120 kgm/s
d 60 kgm/s
HALPLPLPPLL
Answer:
d. 60
Explanation:
If a body of mass 2 kg is moving with a velocity of 30 m/s, then
on doubling its velocity the momentum becomes
a 30 kgm/s
b 90 kgm/s
C 120 kgm/s
d 60 kgm/s
HALPLPLPPLL
Answer:
120
Explanation:
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