Answer:
100 ÷ 9.58 = 10.44 (approximate answer)
The standard deviation of Eric’s data is 0.8°C. Martha conducted the same experiment. Her average temperature was 35.1 with a standard deviation of 1.2°C. Her data are more precise than Eric's
Answer:
less precise than
Explanation:
Answer:
less precise than
Explanation:
its right on edge2020
You and a friend are doing the laundry when you unload the dryer and the discussion comes around to static electricity. Your friend wants to get some idea of the amount of charge that causes static cling. You immediately take two empty soda cans, which each have a mass of 120 grams, from the recycling bin. You tie the cans to the two ends of a string (one to each end) and hang the center of the string over a nail sticking out of the wall. Each can now hangs straight down 30 cm from the nail. You take your flannel shirt from the dryer and touch it to the cans, which are touching each other. The cans move apart until they hang stationary at an angle of 10 degrees from the vertical.
Assuming that there are equal amounts of charge on each can, you now calculate the amount of charge transferred from your shirt.
Answer:
Q_total = 2 10⁻⁶ C
Explanation:
Let's apply the conditions of static equilibrium to this case, in the adjoint we can see a diagram of the forces.
X axis
Tₓ - Fe = 0
Tₓ = Fe
Y axis
[tex]T_{y}[/tex] - W = 0
T_{y} = W
Let's use trigonometry to find the stress components, the angle is measured with respect to the vertical
sin 10 = Tₓ / T
cos 10 = T_{y} / T
Tₓ = T sin 10
T_{y} = T cos 10
we substitute
T sin 10 = Fe
T cos 10 = W
T = mg / cos10
T = 0.120 9.8 / cos 10
T = 1,194 N
Fe = 1,194 sin 10
Fe = 0.2073 N
the electric force is
F = k q₁q₂ / r²
in this case, as the cans touch, they have the same charge and the distance r is searched for by trigonometry
sin 10 = y / L
y = L sin 10
y = 0.30 sin 10
y = 0.052 m
this is the distance from the vertical to one can the distance between the two cans is
y_total = 2y
y_totlal = 2 0.052 = 0.104 m
Fe = F = k q² / r²
q = √ (F r² / k)
q = √ (0.2073 0.104²/9 10⁹) = √ (24.913 10⁻¹⁴)
q = 4.99 10⁻⁷ C
This is the charge of a can, as the transfer is carried out by contact, the flannel transfers half of its charge to the cans and these when separating face one keeps half of the transferred charge, therefore the total charge of the flannel is
Q_total = 4 q
Q_total = 19.97 10⁻⁷ C
Q_total = 2 10⁻⁶ C
A boy throws a ball on a spring scales which oscillates about the equilibrium position with a period of T = 0.5 sec. The amplitude of the vibration is A = 2 cm. Assume the ball does not bounce from the scales’s surface afterwards. Assume the vibration of the scale is expressed mathematically as x(t) = Acos(t + ). Find:
Complete Question
A boy throws a ball on a spring scales which oscillates about the equilibrium position with a period of T = 0.5 sec. The amplitude of the vibration is A = 2 cm. Assume the ball does not bounce from the scales’s surface afterwards. Assume the vibration of the scale is expressed mathematically as x(t) = Acos(t + ). Find:
a) frequency
b) the maximum acceleration
c) the maximum velocity
Answer:
a
[tex]f = 2 \ Hz[/tex]
b
[tex]a_{max} = 3.2 \ m/s^2[/tex]
c
[tex]v_{max} = 0.25 \ m/s[/tex]
Explanation:
From the question we are told that
The period is [tex]T = 0.5 \ sec[/tex]
The amplitude is [tex]A = 2 \ cm = 0.02 \ m[/tex]
The vibration of the scale is [tex]Acos(wt + \phi )[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{1}{T}[/tex]
=> [tex]f = \frac{1}{0.5}[/tex]
=> [tex]f = 2 \ Hz[/tex]
The maximum acceleration is mathematically represented as
[tex]a_{max} = A *(2 \pi f)^2[/tex]
=> [tex]0.02 * (2 * 3.142 * 2)^2[/tex]
=> [tex]a_{max} = 3.2 \ m/s^2[/tex]
The maximum velocity is mathematically represented as
[tex]v_{max} = A * (2 \pi f)[/tex]
=> [tex]v_{max} = 0.02 * (2 * 3.142 * 2)[/tex]
=> [tex]v_{max} = 0.25 \ m/s[/tex]
A truck has a mass of 3000 kg and a velocity
of 10 m/s. Calculate momentum!
Answer:
[tex] \huge{ \boxed{ \sf{30000 \: kg \: {m/s}^{2}}}}[/tex]
Explanation:
[tex] \text{ \underline{ Given}} : [/tex]
[tex] \star[/tex] Mass of a truck ( m ) = 3000 kg
[tex] \star[/tex] Velocity of a truck ( v ) = 10 m / s
Finding the momentum :
[tex] \boxed{ \sf{momentum \: ( \: p \: ) \: = \: mass \: ∗ \: \: velocity}}[/tex]
[tex] \hookrightarrow{ \sf{momentum \: ( \: p \: ) \: = 3000 \: ∗ \: 10}}[/tex]
[tex] \hookrightarrow{ \sf{ \: momentum \: ( \: p \: ) \: = 30000 \: kg \: m/ {s}^{2} }}[/tex]
Hope I helped!
Best wishes ! :D
~[tex] \sf{TheAnimeGirl}[/tex]
Given:-
Mass of the body (m) = 3000 kgVelocity (v) = 10 m/sTo calculate: Momentum of the body.
We know,
p = mv
where,
p = Momentum,m = Mass &v = Velocity.Thus,
p = (3000 kg)(10 m/s)
→ p = 30000 kg m/s (Ans.)
What does the statement "all motion is relative"
mean?
The SI system uses three base units. Question 6 options: True False
Answer:
The answer is false
Explanation:
Though the mostly used SI unit of measurement or the most popular units are the
Length,
Time and
Mass
i.e meter (m), seconds (s), kilogram (kg)
Aside all the above stated units for measurements there are other four basic units which are itemized bellow.
they are
1. Amount of substance - mole (mole)
2. Electric current - ampere (A)
3. Temperature - kelvin (K)
4. Luminous intensity - candela (cd)
A speeder passes a parked police car with a speed of 65 km/h in a 50km/h zone. One second after the speeder has passed the police car, the police begin his pursuit. The police car accelerates with constant acceleration of 2m/s^2.Required:a. How long does it take for the police to catch the speeding car? b. How far did the police car travel before police caught up with the speeder? c. What is the speed of the police car when catches up with the speeder?
Answer:
a) The police will take 18.056 seconds to catch the speedy car, b) The police will travel 326.019 meters before catching the speedy car, c) The speed of the police car when catches up with the speeder is 36.112 meters per second.
Explanation:
Let suppose that speeder moves in a uniform motion, whereas police car has an uniformly accelerated motion.
a) How long does it take for the police to catch the speeding car:
Kinematic equation of each vehicle's position are described:
Speeder
[tex]s_{A} = s_{A,o}+v_{A}\cdot t[/tex]
Police Car
[tex]s_{B} = s_{B,o}+v_{B,o}\cdot t + \frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]
If [tex]s_{A} = s_{B}[/tex], [tex]s_{A,o} = s_{B,o}[/tex], [tex]v_{A} = 18.056\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the resulting expression is done:
[tex]v_{A} \cdot t = v_{B,o}\cdot t +\frac{1}{2}\cdot a_{B}\cdot t^{2}[/tex]
[tex]\frac{1}{2}\cdot a_{B}\cdot t^{2}+(v_{B,o}-v_{A})\cdot t = 0[/tex]
[tex]t \cdot \left(\frac{1}{2}\cdot a_{B}\cdot t +v_{B,o}-v_{A} \right)= 0[/tex]
[tex]t = 0\,s\,\wedge\, t = \frac{2\cdot (v_{A}-v_{B,o})}{a_{B}}[/tex]
[tex]t = \frac{2\cdot \left(18.056\,\frac{m}{s}-0\,\frac{m}{s} \right)}{2\,\frac{m}{s^{2}} }[/tex]
[tex]t = 18.056\,s[/tex]
The police will take 18.056 seconds to catch the speedy car.
b) How far did the police car travel before police caught up with the speeder?
The distance travelled by the police is: ([tex]s_{B,o} = 0\,m[/tex], [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]t = 18.056\,s[/tex])
[tex]s_{B} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (18.056\,s)+\frac{1}{2}\cdot \left(2\,\frac{m}{s^{2}} \right) \cdot (18.056\,s)^{2}[/tex]
[tex]s_{B} = 326.019\,m[/tex]
The police will travel 326.019 meters before catching the speedy car.
c) What is the speed of the police car when catches up with the speeder?
The speed of the police car is represented by the following formula:
[tex]v_{B} = v_{B,o} + a_{B}\cdot t[/tex]
Where [tex]v_{B}[/tex] is the speed of the police car, measured in meters per second.
Given that [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], [tex]t = 18.056\,s[/tex] and [tex]a_{B} = 2\,\frac{m}{s^{2}}[/tex], the final speed of the police car when catches up with the speeder is:
[tex]v_{B} = 0\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right)\cdot (18.056\,s)[/tex]
[tex]v_{B} = 36.112\,\frac{m}{s}[/tex]
The speed of the police car when catches up with the speeder is 36.112 meters per second.
explain how ozone in the atmosphere affects visible light on earth
Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in micro Amps to one decimal place) that your embedded processor can have if you want the battery to last 23 years without being replaced
Answer:
85.12 μAmp
Explanation:
The battery power output = 17.15 Amp-hr
If the battery is to last 23 years, we have to calculate how many hours there are in 23 years
in one year there are 24 hours x 365 day = 8760 hrs
in 23 years there are 23 x 8760 = 201480 hours
maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = 85.12 μA
A golfer needs to sink an 8m putt. She hits the ball, giving it an initial speed of 1.6 m/s, but it stops 1.5m short of the hole. c. What initial speed did the ball need to just make the putt? Assume the acceleration of the grass is constant. Hint: to just make the putt, the speed of the ball will be zero right at the hole. d. In order to make the 1.5m putt, what initial speed does she need to give the ball?
Answer:
1.78 m/s
Explanation:
Distance to putt = 8 m
Distance the ball stops to the putt = 1.5 m
therefore distance traveled by the ball = 8 - 1.5 = 6.5 m
The ball stops at this point 1.5 m from the putt, therefore its final velocity at this point = 0 m/s
the ball was struck with an initial velocity of 1.6 m/s
Using the equation
[tex]v^2[/tex] = [tex]u^2[/tex] + 2as
where v is the final velocity of the ball = 0 m/s
u is the initial speed of the ball = 1.6 m/s
a is the acceleration of the grass
s is the distance the ball travels = 6.5 m
substituting values, we have
[tex]0^2[/tex] = [tex]1.6^2[/tex] + 2(a x 6.5)
0 = 2.56 + 13a
13a = -2.56
a = -2.56/13 = -0.197 m/s^2
If this acceleration of the grass is assumed to be constant, then to the initial speed needed to make the putt will be calculated from
[tex]v^2[/tex] = [tex]u^2[/tex] + 2as
where
v is the final speed at the putt = 0 m/s
u is the initial speed with which the ball is struck = ?
a is the acceleration of the grass = -0.197 m/s^2
s is the distance to the putt = 8 m
substituting values, we have
[tex]0^2[/tex] = [tex]u^2[/tex] + 2(-0.197 x 8)
0 = [tex]u^2[/tex] - 3.152
[tex]u^2[/tex] = 3.152
u = [tex]\sqrt{3.152}[/tex] = 1.78 m/s
Caroline, a piano tuner, suspects that a piano's B4 key is out of tune. Normally, she would play the key along with her B4 tuning fork and tune the piano to match, but her B4 tuning fork is missing! Instead, she plays the errant key along with her A4 tuning fork (which has a frequency of 440.0 Hz), displays the resulting waveform on a handheld oscilloscope, and measures a beat frequency of 15.9 Hz. Then, she plays the errant key along with her C5 tuning fork (which has a frequency of 523.3 Hz) and measures a beat frequency of 67.4 Hz. What frequency is being played by the out-of-tune key? If the B4 key is supposed to produce a frequency of 493.9, is the frequency of the key lower than it should be ("flat") or higher than it should be ("sharp")?
Answer:
455.9 Hz, Flat
Explanation:
The beat frequency is basically the difference between the frequency of the B4 note and the frequency of the A4 tuning fork. This means the B4 note is 15.9 Hz off of 440Hz, and 67.4 Hz off of 525.3 Hz. As B4 is between the two notes, it would make sense to find its frequency by adding 15.9 to 440 and subtracting 67.4 from 523.3, both if which give us a frequency 455.9 Hz for the B4 key. This is because the note doesn't change for the different turning forks so both differences should result in the same frequency. Because the note should be 493.9 frequency but instead has a frequency of 455.9 Hz, it is flat because the frequency is lower than it is supposed to be.
Hope this helped!
This question deals with the phenomenon of "Beat Frequency".
The frequency being played by the B4 key is "455.9 Hz", which is "lower (flat) than it should be".
When two waves with a slightly different frequency interfere with each other, they produce a pattern known as beat frequency. The beat frequency is the difference between the frequencies of the interfering wave. Since the frequency of B4 key produces beat frequency with both the C5 tuning fork (which has a frequency of 523.3 Hz) and the A4 (which has a frequency of 440 Hz). Hence, the frequency of the B4 key must be between the range of these frequencies, that is, 440 Hz to 523.3 Hz.
ANALYZING THE BEAT FREQUENCY WITH A4 TUNING FORK:
Beat Frequency with A4 tuning fork = Frequency of B4 - Frequency of A4
Frequency of B4 = Frequency of A4 + Beat Frequency with A4 tuning fork
Frequency of B4 = 440 Hz + 15.9 Hz
Frequency of B4 = 455.9 Hz
ANALYZING THE BEAT FREQUENCY WITH C5 TUNING FORK:
Beat Frequency with C5 tuning fork = Frequency of C5 - Frequency of C5
Frequency of B4 = Frequency of C5 - Beat Frequency with C5 tuning fork
Frequency of B4 = 440 Hz - 67.4 Hz
Frequency of B4 = 455.9 Hz
Both the answers validate each other.
Since the expected frequency of the B4 key is 493.9 Hz, which is greater than the actual frequency produced by the B4 key. Hence, we can say that "the frequency of the key lower than it should be ("flat")"
The attached picture shows the general formula of the beat frequency.
Learn more about beat frequency here:
https://brainly.com/question/10703578?referrer=searchResults
Which is NOT one of the 11 organ systems of the body?
Integumentary system
Immune system
Skeletal system
Digestive system
Answer:
Immune
Explanation:
because I remember taking that test and I I specifically remember that question and that's the answer
A 2250 kg truck has a velocity of 25 m/s to the east. What is the momentum of the truck? p = mv
The momentum of the truck is "56250 kgm/s".
According to the question,
Mass,
m = 2250 kgVelocity,
v = 25 m/sAs we know,
→ [tex]p = mv[/tex]
By putting the values, we get
→ [tex]= 2250\times 25[/tex]
→ [tex]= 56250 \ kgm/s[/tex]
Thus the answer is right.
Learn more:
https://brainly.com/question/21433514
Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground.
A) At a home 5.50km away from the antenna, what average pressure does this wave exert on a totally reflecting surface? answer in PA units
B) At a home 5.50km away from the antenna, what are the amplitudes of the electric and magnetic fields of the wave? answer in N/C, T
C) At a home 5.50km away from the antenna, what is the average density of the energy this wave carries? answer in J/m^3
D) For the energy density in part (c), what percentage is due to the electric field?
E) For the energy density in part (c), what percentage is due to the magnetic field?
Answer:
A. Using
I= P/ A
So = P/ 2π R²
So=777E3 watt/ 2*πx (²5.5E3m)
4.3*10^ -3W/m²
So pressure= 2I/c
=2* 4.3*10^ -3W/m²/ 3*10^ 8
2.7*10^-11p.a
B.
Using Emax= √2I/Eoc
So
√ 2*4.3*10^-3/8.8E-12*3*10^8
= 5.48V/m
So
Bmax= Emax/c
5.58/3E8
= 1.83*10-9T
D.average density is given by
= Eo(Emax/√2)²
= 8.85E-12(5.48/√2)²
1.3*10^10W/m²
The second law of thermodynamics leads us to conclude what?
a. the total energy of the universe is constant
b. disorder in the universe is increasing with the passage of time
c. it is theoretically impossible to convert work into heat with 100% efficiency
d. the total energy in the universe is increasing with time
e. the total energy in the universe is decreasing with time
Answer:
b. disorder in the universe is increasing with the passage of time
Explanation:
The second law of thermodynamics states that the total entropy of an isolated system increases over time . The entropy of a system measures the randomness or disorder in the system . So the we can state, in terms of disorder , the second law of thermodynamics as follows .
The disorder in the universe is increasing with the passage of time .
The second law of thermodynamics leads us to conclude that disorder in the universe is increasing with the passage of time. So, here, option b is the correct option.
Entropy is a measure of the disorder or randomness in a system. The second law of thermodynamics states that in any spontaneous process, the total entropy of an isolated system (including both the system and its surroundings) tends to increase or, at best, remain constant. This means that over time, natural processes tend to lead to an increase in the overall disorder or randomness of the system. It's important to note that while the increase in entropy is a statistical tendency on a macroscopic scale, there can still be localized decreases in entropy within a system.
Learn more about the law of thermodynamics here.
https://brainly.com/question/17405303
#SPJ6
The pilot of a helicopter drops a lead brick from a height of 500 m. (assume no air resistance and g~10 m/s2) roles Evalu How long does it take to reach the ground?
Answer:
The value is [tex]t = 10 \ s[/tex]
Explanation:
From the question we are told that
The height is [tex]h = 500 \ m[/tex]
The acceleration is [tex]g = 10 \ m/s[/tex]
Generally from the second equation of motion we have that
[tex]h = ut + \frac{1}{2} a t^2[/tex]
Here the initial velocity u = 0 m/s given that the stone started from rest
So
[tex]500 = 0 (t) +0.5 * (10) t^2[/tex]
=> [tex]t = 10 \ s[/tex]
A car moves in a straight line at a speed of 72.1 km/h.
How far (in km) will the car move in 7.74 minutes at this speed?
Speed=distance/time.
7.74mins=464.4secs.
72.1=distance/464.4.
distance=464.4×7.21.
=3348.324m.
Calculate the change in momentum of a 0.5kg ball that strikes the floor at 15 m/s and bounces back up at 12 m/s
Answer:
The change in momentum is: [tex]13.5\,\frac{kg\,m}{s}[/tex]
Explanation:
Let's define that vectors pointing up are positive, and vectors pointing down negative.
Then we express the initial momentum of the ball as the product of its mass times the velocity, and include the negative sign since this momentum is pointing down (velocity vector is pointing down):
[tex]P_i=-0.5\,(15) \frac{kg\,m}{s} =-7.5\, \frac{kg\,m}{s}[/tex]
the final momentum is positive (pointing up) and given by the product:
[tex]P_f=0.5\,(12) \frac{kg\,m}{s} =6\, \frac{kg\,m}{s}[/tex]
Therefore, since the change in momentum is defined as the difference between the final momentum minus the initial one, we get:
[tex]P_f-P_i=6-(-7.5)\,\frac{kg\,m}{s} =13.5\,\frac{kg\,m}{s}[/tex]
If three resistors 2ohm, 3ohm and 4ohm are connected in a circuit. Calculate the equivalent resistance of the combination
You haven't described whether they're connected in series or in parallel. Actually, there are eight (8) different ways they can be arranged, and each way has a different equivalent resistance.
==> All 3 resistors in series. Equivalent resistance = (2+3+4) = 9.000 ohms
==> All 3 resistors in parallel.
Equivalent resistance = 1 / (1/2 + 1/3 + 1/4) = (12/13) ohm or 0.923 ohm
==> The 2 ohm resistor in series with (the 3 and the 4 in parallel)
Equivalent resistance = 2 + 1/(1/3 + 1/4) = 3.714 ohms
==> The 3 ohm resistor in series with (the 2 and the 4 in parallel)
Equivalent resistance = 3 + 1/(1/2 + 1/4) = 4.333 ohms
==> The 4 ohm resistor in series with (the 2 and the 3 in parallel)
Equivalent resistance = 4 + 1/(1/2 + 1/3) = 5.200 ohms
==> The 2 ohm resistor in parallel with (the 3 and the 4 in series)
Equivalent resistance = 1.556 ohms
==> The 3 ohm resistor in parallel with (the 2 and the 4 in series)
Equivalent resistance = 2.000 ohms
==> The 4 ohm resistor in parallel with (the 2 and the 3 in series)
Equivalent resistance = 2.222 ohms
Andy took a bus and then walked from his home to downtown.
For the first 1.6 hour, the bus drove at an average speed of 15
km/h. For the next 0.4 hours, he walked at an average speed
of 4.5 km/h. What was the average speed for the whole
journey?
Answer:
12.9
Explanation:
1.6x15=24
0.4x4.5=1.8
24+1.8=25.8
1.6+0.4=2
25.8/2=12.9
A toy car that is 0.12 m long is used to model the actions of an actual car that
is 6 m long. Which ratio shows the relationship between the sizes of the
model and the actual car?
A: 5:1
B: 1:50
C: 50:1
D: 1:5
Answer:
it is B 1:50
Explanation:
just did it on apex
What is the final velocity of a 1400 kg drag racer that’s engine applies 910000 N of Force for 120m
Answer:
468,000,000 m/sExplanation:
According to newtons first law of motion;
Force = mass * acceleration
F = ma
where a = v-u/t
v is the final velocity
u is the initial velocity
t is the time taken
F = m(v-u/t)
Ft = m(v-u)
Given parameters;
F = 910,000N
t = 120m = 120 *60 = 7200secs
m = 1400kg
u = 0m/s
Required parameter
Final velocity v
Substituting the given parameters into the formula to get v;
Ft = m(v-u)
910,000 * 72000 = 1400(v-0)
910,000 * 72000 = 1400v
v = (910,000 * 72000)/1400
v = 910,000 * 720/14
v = 910,000 * 60
v = 468,000,000 m/s
Hence the final velocity of the drag racer is 468,000,000 m/s
Challenge Problem
5.
A man is standing on the top of a hill and sees a flagpole he knows is 45 feet high. The angle of
depression to the bottom of the pole is 12 degrees, and the angle of elevation to the top of the
pole is 16 degrees. Find his distance from the pole.
Answer: his distance from the pole is 90.16 ft
Explanation:
based on the diagram of the question i will upload along this answer;
45 - d is equal to part of the pole below the horizontal line
d/x = tan(16) ; x = d/tan(16)............equ1
(45-d) / x = tan(12); x = (45-d) / tan(12)-----------equ2
∴ d/tan(16) = (45-d) / tan(12)
d.tan(12) = (45-d).tan(16)
d.tan12 = (45×tan16) - d.tan16
0.2125d = 12.9034 - 0.2867d
0.2125d + 0.2867d = 12.9034
0.4992d = 12.9034
d = 12.9034/0.4992
d = 25.85 ft
now substitute value of d into any of our previous equation, lets take equation 1
x = 25.85 / tan(16)
x = 25.85 / 0.2867
x = 90.16 ft
Therefore his distance from the pole is 90.16 ft
Plz help me :( The average human can run 60.35 meters in 9 seconds, calculate the average speed in meters per second (m/s)
Answer:
6.705555556 m/s
Explanation:
60.35 divide by 9
True or False: If a point charge has electric field lines that point into it, the charge must be positive.
Answer:
False
Explanation:
If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I = 1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Answer:
0.00004666N
Explanation:
We know that
intensity (I) = P/ A
Where
P= power
A= Area
So lets say that the power absorbed
Will be = Intensity x Area
Which Is = 1.4 x 10^3 x(10)
So
14000 Watt = 14 kWatt
However we know that radiation pressure is equal to
time-averaged intensity all over the speed of light in free space
So
P = (1.4 x 1000)/c
But
F= P x A
So
((1.4 x 1000)/(3 x1 0^8)) x 10
Which is
=0.000046666N
Explanation:
In an evironmental system of subsystem, the mass balance equation is:__________.
Answer:
Explanation:
The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as
Input = Output + Accumulation
The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.
Which is the least dense liquid?
Answer:
lamp oil.
Explanation: the ping pong ball doesnt sink through the lamp oil.
A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest point, which statement is true? (A) Its velocity is not zero, but its acceleration is zero. (B) Its velocity and its acceleration are both zero. (C) Its velocity is perpendicular to its acceleration. (D) Its acceleration depends on the angle at which the ball was thrown. (E) None of the above statements are true.
Answer:
C. Its velocity is perpendicular to its acceleration
Explanation:
Because acceleration is always perpendicular to the velocity when the velocity will change direction without change it's magnitude
What must the charge (sign and magnitude) of a particle of mass 1.50 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?
Answer:
q = -2.19 x 10⁻⁵ C
Explanation:
Given;
mass of the particle, m = 1.5 g = 0.0015 kg
magnitude of electric field, E = 670 N/C
Electric field is given by;
[tex]E = \frac{F}{q}[/tex]
where;
q is the magnitude of the
f is the force of the charge
f = mg
[tex]E = \frac{F}{q}\\\\E = \frac{mg}{q}\\\\q = \frac{mg}{E}\\\\q = \frac{0.0015*9.8}{670} \\\\q = 2.19*10^{-5} \ C[/tex]
Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative
q = -2.19 x 10⁻⁵ C