This question is incomplete, the complete question is;
In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.
What is the effective diameter of the circular laser aperture at the Maui ground station
Answer:
the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
Explanation:
Given the data in the question;
Separation between observer and point L = 354 km = 354000 m
Linear separation D = 9.1 m
wavelength λ = 500 nm = 500 × 10⁻⁹ m
Now, for small angles;
θ = D / L
θ = 9.1 m / 354000 m
θ = 2.57 × 10⁻⁵ rad
For a circular aperture;
sinθ = ( 1.22 × λ ) / d
for small angles;
θ = ( 1.22 × λ ) / d
so
θ = 2 × θ
θ = 2 × [( 1.22 × λ ) / d]
we substitute
2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]
2.57 × 10⁻⁵ = 0.00000122 / d
d = 0.00000122 / 2.57 × 10⁻⁵
d = 0.04747 m
d = ( 0.04747 × 100 )m
d = 4.747 cm
Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm
The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2 pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2
Answer:
Explanation:
a)
Given that:
V = 25 mi/hr
To ft/sec, we have:
[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{110}{3} ft/s[/tex]
[tex]\rho = 0.075 \ lb/ft^3[/tex]
[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]
[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]
[tex]C_d = 0.28[/tex]
A = 25ft²
Recall that:
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]
[tex]F_d =10.967 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]
For 70 miles per hour, we have:
[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{308}{3} ft/s[/tex]
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]
[tex]F_d =85.99 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]
A train is moving with a speed of 100 m/s. If the train is traveling south, at what position will it be 3 minutes after passing the +1,000-meter position marker ?
Remember, south is the negative direction and when you use the time, it must be in units of seconds. You will be applying one of the equations from above to solve this problem. And you must include a + sign if the final position is positive or a - sign if the final position is negative.
Answer:
The position after 3 minutes is - 800 m.
Explanation:
speed, v = 100 m/s
time, t = 3min = 180 s
initial position, x = 1000 m
let the distance traveled in 3 minutes is d
d = 100 x 180 = - 18000 m
So, the position is
= - 18000 + 1000 = - 800 m
Now imagine that a beam of moving electrons enters the quadrupole, with velocity parallel to the charged rods. When the beam enters the device, its cross-sectional profile is circular. What will the beam look like when it exits the quadrupole on the other side
Answer:
The beam of electrons would be flattened into a oval that is long in the x-axis and short in the y-axis.
Explanation:
A beam of moving electrons enters the quadrupole, with velocity parallel to the charged rods. When the beam enters the device, its cross-sectional profile is circular. On the other side, the beam of electrons would be flattened into a oval that is long in the x-axis and short in the y-axis.
The directions is shown in figure.
Imagine two circular plates; one is solid and the other has a hole cut out of the center. Both plates have the same radius, same thickness, and same mass. The same force F is applied tangential to the edge of each plate in such a way that the plates rotate about an axis passing through the center and perpendicular to the surface of the plates.
Which one of the following statements is true regarding the angular acceleration?
1. Both plates will rotate with the same angular acceleration.
2. The solid plate will have the greater angular acceleration.
3. The plate with the hole will have the greater angular acceleration.
Which of the following statements helps to explain the question asked above? (Select all that apply.)
1. Because both plates have the same mass, they will have the same moment of inertia.
2. Angular acceleration is inversely proportional to the moment of inertia.
3. Angular acceleration is directly proportional to the moment of inertia.
4. The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia.
5. Both plates will be subjected to the same torque.
Answer:
the correct statement is 2. The solid plate will have the greater angular acceleration.
the correct phrase is 4. The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia.
Explanation:
Newton's second law expression for rotational motion is
τ = I α (1)
where the torque is
τ = F r
in this case, as the discs have the same radius and the applied force is the same, the torque is the same on the two discs.
The moment of inertia is given by the expression
I =∫ r² dm
for bodies with high symmetry are tabulated
the moment of inertia for in disk solid is I₁ = ½ m R₂²
the moment for a disk with a hole I₂ = ½ m (R₁² + R₂²)
We can see that the moment of inertia of the disk with the hole is greater than the moment of inertia of the solid disk.
Let's use equation 1
α = τ/I
therefore the angular acceleration is lower for the body with the higher moment of inertia, consequently the solid disk has higher angular acceleration
the correct statement is 2
The reason is because the moment of inertia is higher for the hollow disk.
the correct phrase is 4
A 1.5 kg rock is dropped from a height of 2.0 meters onto a spring that
compresses and brings the rock to rest. (Assume no losses to thermal
energy.) How much energy is in the system before the drop.
Answer:
29.4 J
Explanation:
Before the drop, the system has only the gravitational potential energy, and this energy us given by mass×gravity×height:
1.5•9.8•2 = 29.4 J
20 points, im begging for help‼️
How much capacitance is needed to
store 0.00100 J of energy when the
charge on the capacitor is
4.86 x 10-5 C?
[?] x 10?!F
Answer:
Capacitance= 1.18×10^-6
Answer: 1.18*10^-6
Explanation:
A box rests on a frozen pond, which serves as a frictionless horizontal surfaceIf a fisherman applies a horizontal force with magnitude 51.0 N to the box and produces an acceleration of magnitude 3.00 m/s2 , what is the mass of the box?
Answer:
the mass of the box is 17 kg
Explanation:
Given;
magnitude of the force applied by the fisherman, F = 51 N
magnitude of acceleration of the box, a = 3 m/s²
the mass of the box is calculated using Newton's law of motion;
F = ma
where;
m is the mass of the box
m = F / a
m = (51) / (3)
m = 17 kg
Therefore, the mass of the box is 17 kg
Three ice skaters, numbered 1, 2, and 3, stand in a line, each with her hands on the shoulders of the skater in front. Skater 3, at the rear, pushes forward on skater 2. Assume the ice is frictionless.
Identify all the action/reaction pairs of forces between the three skaters. Draw a free-body diagram for skater 2, in the middle.
Answer:
Following are the responses to the given question:
Explanation:
The mass (mg) is downwards, the typical [tex]N_2[/tex] pressure is upward. This pushing force of [tex]\vec{F}_{3 \to 2}[/tex] is pushed forward by skater 3 on skater 2. (considered as rightward direction). The strength of the slater 2 in reply is the slater [tex]\vec{F}_{2\to 3}[/tex] Skater three moves to the left.
Its push force [tex]\vec{F}_{2\to1}[/tex] imparted to the skateboarder 2 in relation those above forces The force[tex]\vec{F}_ {1\to 2}[/tex] on skater 2 by skater 1 is as a response, to a left. Skater 2's free body system is as follows:
how many protons does a neutral atom of oxygen-16 have
Answer:
eight
Explanation:
it's atomic number is 8 which mean that an oxygen atom has eight protons in it's nucleus
A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?
Answer:
3.73 * 10^16 photons/sec
Explanation:
power supply = 3.0 V
Emits 440 nm blue light
current in LED = 11 mA
efficiency of LED = 51%
Calculate the number of photons per second the LED will emit
first step : calculate the energy of the Photon
E = hc / λ
=( 6.62 * 10^-34 * 3 * 10^8 ) / 440 * 10^-9
= 0.0451 * 10^-17 J
Next :
Number of Photon =( power supply * efficiency * current ) / energy of photon
= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17
= 3.73 * 10^16 photons/sec
What is the primary function of the lower
respiratory system?
O to move nutrients to the cells throughout the
body
O to move blood to the cells throughout the body
O to extract oxygen from the air that the body breathes in
O to extract oxygen from the blood that the body makes
Answer:
C
Explanation:
100%
An unstretched ideal spring hangs vertically from a fixed support. A 0.4 kg object is then attached to the lower end of the spring. The object is pulled down to a distance of 0.35 m below the unstretched position and released from rest at time t= 0. A graph of the subsequent vertical position y of the lower end of the spring as a function of t is given above, where y= 0 when the spring was initially unstretched. At which time is the upward velocity of the object the greatest?
Answer:
The correct answer will be "0.25 sec".
Explanation:
The graph of the given question is attached below.
According to the graph of the question,
Time,
T = 1 sec
For the upward velocity,
⇒ [tex]t = \frac{T}{4}[/tex]
By putting the value, we get
⇒ [tex]=\frac{1}{4}[/tex]
⇒ [tex]=0.25 \ sec[/tex]
A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a) S = 63.2 km/h
b) V = 63.2 km/h*(-0.316 , 0.949)
Explanation:
Let's define:
North as the positive y-axis
East as the positive x-axis.
Also, remember the relation:
Distance = Time*Speed
Let's assume that she starts at the position (0km, 0km)
Then she travels due North at 90km/h for two hours, then the displacement is
90km/h*2h = 180km to the north
Then the new position is:
(0km, 180km)
Then she travels West at 60km/h for one hour.
Then the distance traveled to the West (negative x-axis) is:
60km/h*1h = 60km to the west
Then the new position is:
(-60km, 180km).
a) The average speed is defined as the quotient between the displacement and the time.
We know that the total time traveled is 3 hours.
And the displacement is the difference between the final position and the initial position.
this is:
D = √( -60km - 0km)^2 + (180km - 0km)^2)=
D = √( (60km)^2 + (180km)^2) = 189.7 km
Then the average speed is:
S = (189.7 km)/(3 h) = 63.2 km/h
b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.
So, if the final position is (-60km, 180km)
We need to find a vector that represents the same angle, but that is on the unit circle.
Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:
(-60km/ 189.7 km, 180km/ 189.7 km)
(-60/189.7 , 180/189.7)
We can just check that the module of the above versor is 1.
[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]
Then the average velocity is:
V = 63.2 km/h*(-60/189.7 , 180/189.7)
We can simplify our versor so the velocity equation is easier to read:
V = 63.2 km/h*(-0.316 , 0.949)
Unpolarized light of intensity 0.0288 W/m2 is incident on a single polarizing sheet. What is the rms value of the electric field component transmitted
Answer:
the rms value of the electric field component transmitted is 3.295 V/m
Explanation:
Given;
intensity of the unpolarized light, I = 0.0288 W/m²
For unpolarized light, the relationship between the amplitude electric field and intensity is given as;
[tex]E_{max} = \sqrt{2\mu_0cI} \\\\E_{max} = \sqrt{2(4\pi \times 10^{-7})(3\times 10^8)(0.0288)} \\\\E_{max} = 4.66 \ V/m[/tex]
The relationship between the rms value of the electric field and the amplitude electric field is given as;
[tex]E_{rms} = \frac{E_0}{\sqrt{2} } =\frac{E_{max}}{\sqrt{2} } \\\\E_{rms} = \frac{4.66}{\sqrt{2} }\\\\E_{rms} = 3.295 \ V/m[/tex]
Therefore, the rms value of the electric field component transmitted is 3.295 V/m
A spinning satellite begins to unfold two solar panels as shown. As the
panels extend from the satellite, what is the result?
Its angular momentum of a satellite will drop even as satellite panel stretch because of the reduction in angular velocity brought on by the drag effect.
Why then do stars spin more quickly during their demise?A portion of the star's mass is blown off during the supernovae that come before the creation of black holes, taking some of the star's total angular momentum with it.The leftover material sinks into the star's core while continuing to spin rapidly.
What is the name of the spinning effect?Magnus effect: When there is relative movement between both the spinning item and the fluid, a sideways force is generated on a rotating cylindrical or spherical object immersed in the fluid (liquid or gas).In honor of the German chemist and physicist H.G.
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In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 70 kg, are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N.
Required:
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Answer:
21420 J
Explanation:
Given that:
mass of the rider = 70 kg
the tension of the rope = 350 N
Using the concept of conservation of energy;
Work done = change in the Total energy ( ΔT.E)
where;
Work done (W) = ΔK.E + ΔP.E + ΔThermal Energy
Recall that the man proceeds with a constant speed, thus the change in K.E energy will be zero.
As such:
W = ΔP.E + ΔThermal Energy
We can now say that:
The thermal energy = W - ΔP.E
here;
W = force × displacement
The thermal energy = (350 × 120) - (70 × 9.8 × 30)
= 42000 - 20580
= 21420 J
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.
Answer:
Explanation:
From the given information:
The initial PE [tex](PE)_i[/tex] = m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = [tex]P.E_f -P.E_i[/tex]
ΔP.E = 0 - [tex](PE)_i[/tex]
ΔP.E = [tex]-P.E_i[/tex]
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]
[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]
this can be re-written as:
[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]
[tex]\Delta U =(0.70) (490.5)[/tex]
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
Let’s use these equations to compare the electrostatic force and the gravitational force in a few different situations. In each case, calculate the strength of the electrostatic force and the strength of the gravitational force.
Two electrons separated by 1 cm
q = 1.6 x 10-19 C
m = 9.1 x 10-31 kg
d = 1 cm
What is the electric force?
What is the gravitational force?
Which force will dominate the motion of the electrons?
Answer:
Electric force is the attractive force between the electrons and the nucleus. It works the same way for a negative charge, you also have an electric field around it. ... Now, like charges repel each other and opposite charges attract.
The gravitational force is a force that attracts any two objects with mass. We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. ... This is called Newton's Universal Law of Gravitation.
electric force
This table shows clearly that the electric force dominates the motion of electrons in atoms. However, on a macroscopic scale, the gravitational force dominates. Since most macroscopic objects are neutral, they have an equal number of protons and electrons.
Explanation:
What recommendations would you give to the global government to help Decrease the global effects of human impact on the environment mystery recommendations and how they will positively impact our planet
Answer:
We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.
Explanation:
5 ways our governments can confront climate change
PROTECT AND RESTORE KEY ECOSYSTEMS
SUPPORT SMALL AGRICULTURAL PRODUCERS
PROMOTE GREEN ENERGY
COMBAT SHORT-LIVED CLIMATE POLLUTANTS
BET ON ADAPTATION, NOT JUST MITIGATION
Calculating Acceleration
Initial
velocity
Time to travel
0.25 m
Final
velocity
Acceleration
Time to travel
0.50 m
# of
washers
11
(m/s)
V2
(m/s)
ti
(s)
t₂
(s)
a = (v2 - v4)/(t2-tı)
(m/s)
1
0.11
0.28
2.23
3.13
0.19
2
0.13
0.36
1.92
2.61
The acceleration of the car with two washers added to the string would be
I can not even read this question.
What are you trying to even say?
The acceleration of the car with two (2) washers added is equal to 0.33 [tex]m/s^2[/tex].
Given the following data:
Initial velocity = 0.13 m/s.Final velocity = 0.36 m/s.Initial time = 1.92 seconds.Final time = 2.61 seconds.What is an acceleration?An acceleration can be defined as the rate of change of velocity of an object with respect to time and it is measured in meter per seconds square.
How to calculate average acceleration.In Science, the average acceleration of an object is calculated by subtracting its initial velocity from the final velocity and dividing by the change in time for the given interval.
Mathematically, average acceleration is given by this formula:
[tex]a = \frac{V\;-\;U}{t_f-t_i}[/tex]
Where:
V is the final velocity.U is the initial velocity.[tex]t_i[/tex]initial time measured in seconds.[tex]t_f[/tex] final time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]a = \frac{0.36\;-\;0.13}{2.61\;-\;1.92}\\\\a=\frac{0.23}{0.69}[/tex]
a = 0.33 [tex]m/s^2[/tex]
Read more on acceleration here: brainly.com/question/24728358
Which statement accurately represents the arrangement of electrons in Bohr’s atomic model?
Electrons move randomly in the relatively large space surrounding the nucleus.
Electrons move around the nucleus in fixed orbits of equal levels of energy.
Electrons move around the nucleus in fixed orbits of increasing levels of energy.
Electrons vibrate in fixed locations around the nucleus.
Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy
Explanation:
The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.
It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.
Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m and we assume fully developed internal flow, find the pressure drop across this pipe length.
Answer:
[tex]\triangle P=1.95*10^{-4}[/tex]
Explanation:
Mass [tex]m=0.001[/tex]
Diameter [tex]d=1.2m[/tex]
Length [tex]l=10m[/tex]
Generally the equation for Volume flow rate is mathematically given by
[tex]Q=AV[/tex]
[tex]V=\frac{Q}{\pi/4D^2}[/tex]
[tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]
[tex]V=8.84*10^{-4}[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]F=\frac{64}{Re}[/tex]
Where Re
Re=Reynolds Number
[tex]Re=\frac{pVD}{\mu}[/tex]
[tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]
[tex]Re=1040[/tex]
Therefore
[tex]F=\frac{64}{Re}[/tex]
[tex]F=\frac{64}{1040}[/tex]
[tex]F=0.06[/tex]
Generally the equation for Friction factor is mathematically given by
[tex]Head loss=\frac{fLv^2}{2dg}[/tex]
[tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]
[tex]H=19.9*10^{-9}[/tex]
Where
[tex]H=\frac{\triangle P}{\rho g}[/tex]
[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]
[tex]\triangle P=H*\rho g[/tex]
[tex]\triangle P=1.95*10^{-4}[/tex]
Boron is one position to the left of carbon on the periodic table. The atomic number of carbon is 6. Given its position on the periodic table what is the atomic number of boron?
Answer:
5
Explanation:
The diagram shows the molecular structure of ethane. What is the chemical
formula for ethane?
Ethane
H H
H-C-C-H
| |
H H
Vector A= 3.7 i + 1.0 j and vector B = 3.0 i + 6.5 j. What is vector
(A-B).A?
Answer:
(A - B).A = -2.91
Explanation:
First, let's define the sum and dot product of vectors.
For two vectors V = (x₁, y₁) and W = (x₂, y₂) we have:
sum (or subtraction):
V + W = (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂)
dot product:
V.W = (x₁, y₁).(x₂, y₂) = x₁*x₂ + y₁*y₂
Here remember the notation:
V = x₁*i + y₁*j = (x₁, y₁)
Now let's solve our problem, we have:
A = (3.7, 1.0)
B = (3.0, 6.5)
Then:
(A - B).A = ( (3.7, 1.0) - (3.0, 6.5) ).(3.7, 1.0)
= (3.7 - 3.0, 1.0 - 6.5).(3.7, 1.0)
= (0.7, -5.5).(3.7, 1.0) = (0.7*3.7) + (-5.5)*(1.0) = -2.91
The engine of a locomotive exerts a constant force of 6.8 105 N to accelerate a train to 80 km/h. Determine the time (in min) taken for the train of mass 1.1 107 kg to reach this speed from rest.
Answer:
t = 6 minutes
Explanation:
Given that,
Force,[tex]F=6.8\times 10^5\ N[/tex]
Initial speed of the train, u = 0
Final speed of the train, v = 80 km/h = 22.22 m/s
The mass of the train, [tex]m=1.1\times 10^7\ kg[/tex]
We need to find the time taken by the train to come to rest. We know that,
F = ma
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s[/tex]
or
t = 6 minutes (approx)
So, the required time is equal to 6 minutes.
PLS HELP ME 100 POINTS PLS I NEED HELP QUICK PLS
For this project, you are expected to submit the following:
1. Your Student Guide with completed Student Worksheet
2. Your scale model of the solar system
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Conduct research on the actual sizes of the planets.
a) Do research to find the actual sizes of the Sun and the planets. This information is typically represented as diameter in kilometers (km). Recall that diameter is the length of the imaginary straight line from one side of a figure, such as a sphere, to the opposite side of the figure. This line passes through the center of the figure.
b) Record the actual diameters of the Sun and the planets in the first column of the table in the Student Worksheet.
c) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 3: Determine the scaled sizes of the planets.
a) Go to a reliable website to find a solar system model calculator.
b) Decide how big you want the Sun in your model to be. For example, you could assign your Sun to be 300 mm. Input this figure in the calculator, and the calculator will determine the diameters of the eight planets for you. You want to make sure that the Sun is big enough so that the smallest planet will still be big enough to draw.
c) Record information from the calculator in the second column of the table in the Student Worksheet.
d) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 4: Create a scale model of the solar system.
a) Draw and cut construction paper models of the Sun and the planets using the scaled measurements from the table.
b) Glue the models on the poster board. You can glue or tape poster boards together if necessary. Be sure to put the Sun in the center and to put the planets and a drawing of their orbits in order from nearest to farthest from the Sun.
Note: Remember that in this model, the diameter of the planets is scaled but the distance of the planets from the Sun is not. That means your model does not accurately represent the distances of the planets from the Sun so you need not worry about these measurements.
c) Label the Sun and the planets.
d) Put an attention-catching title above or below your model.
e) Write your name on the back of your poster board.
Step 5: Complete the Student Worksheet.
a) Make sure the table in the Student Worksheet is complete.
b) Answer the questions in the Student Worksheet.
c) Check to make sure you added the sources you used for this project in the Student Worksheet.
Step 6: Evaluate your project using this checklist.
If you can check each of the following boxes, you are ready to submit your project.
Did you conduct research to find the actual size of the Sun and the planets? Did you record this information in the table in the Student Worksheet?
Did you use a solar system model calculator to determine the scaled size of the Sun and planets? Did you record this information in the Student Worksheet?
Did you add the links of the websites you used for this project to the Student Worksheet?
Did you use the scaled sizes to create models of the Sun and the planets?
Did you put your model together in a way that represents the solar system (Sun in the center and planets in order from nearest to farthest from the Sun)?
Did you label each component of your model?
Did you add an attention-catching title above or below your model?
Did you write your name on the back of your poster board?
Did you complete the Student Worksheet at the end of this guide?
Step 7: Revise and submit your project.
a) If you were unable to check off all the requirements on the checklist, go back and make sure that your project is complete. Save your project before submitting it.
b) Turn in your scale model of the solar system to your teacher. Be sure that your name is on it.
c) Submit your Student Guide through the virtual classroom.
d) Congratulations! You have completed your project.
Answer
I hope this help....
Explanation:
Answer:
Hope this helps
Explanation:
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
To know more about thermal energy follow
https://brainly.com/question/19666326
A spring, with a spring constant of 4000 N/m, is oriented horizontally, and compressed by 10cm. When released, the spring launches a block of mass 1.0 kg along a 5.0-m horizontal section of track, where the coefficient of friction between the block and track is 0.20. The block then goes up a frictionless ramp angled at 60o with the horizontal. How high up the ramp does the block go before it starts to slide back down
Answer:
[tex]d=1.2m[/tex]
Explanation:
From the question we are told that:
Spring constant [tex]k= 4000 N/m[/tex]
Compressed [tex]l_d= 10cm=>0.10[/tex]
Mass [tex]m=1.0kg[/tex]
Length of horizontal section [tex]l=5.0-m[/tex]
Coefficient of friction [tex]\mu=0.20[/tex]
Angle [tex]\theta=60 \textdegree[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=\mu mgL+mgdsin\theta[/tex]
[tex]\frac{1}{2}k*l_d^2=\mu mgL+mgdsin\theta[/tex]
[tex]\frac{1}{2}(4000)*0.1^2=0.2*1*9.8*5+1*9.8*d*sin60[/tex]
[tex]d=1.2m[/tex]
A 4.00 kg ball is swung in a circle on the edge of a 1.50 m rope. The time it takes for the ball to complete one rotation is 3.40 s. Please show all work and equation.
a) What is the velocity of the ball?
b) What is the acceleration of the ball?
c) What is the force on the ball?
Answer:
The answer is below
Explanation:
The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.
a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:
Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters
The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s
b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):
v = u + at
2.77 = 3.4a
a = 0.82 m/s²
c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N