Answer:
Distance from sun to Neptune = 4.495E9 km
Time for light to travel = 4.495E9 / 3E5 sec = 14,980 sec
That is from sun to Neptune time fof light = 250 min
Time for light to travel from sun to earth is about 8 min
So the time from Neptune would be 242 to 258 min depending on position of Neptune - Note that Neptune is about 30X as far from the sun as earth and
250 min / 8 min is roughly 30
The uniform motion of kinematics allows us to find the time it takes for light to arrive from Neptune to Earth, which varies between:
t₁ = 1.45 10⁴ s and t₂₂= 1.55 10⁴ s
depending on the relative distance of the two planets
given parameters
The speed of light c = 300,000 km / s = 3 10⁸ m / s The distance from Neptune to Sum
to find
The time when light arrives from Neptune to Earth
They velocit of an electromagnetic wave is constant, so we can use the uniform motion relationships
v = d / t
t = d / v
where v is the speed of light, d the distance and y time, in this case the speed of the wave is the speed of light (v = c)
We look in the tables for the distances and the rotation periods around the sun
distance ( m) period (s)
Sun Neptunium 4.50 10¹² 5.2 10⁹
Sun - Earth 1.5 10¹¹ 3.2 10⁷
With the data of the period it is observed that the rotation of Neptune is much greater than that of Eart rotation around the sun, for which we will assume that Neptunium is fixed in space and the Earth may be in its aphelion or perihelion, maximum approach o away distance from the sun, consequently we calculate the time for the two cases:
Maximum approach
positions relative distance from the dos Plantetas is
Δd = [tex]x_{Neptuno - Sum} - x_{Earth - Sum}[/tex]d
Δd = 4.50 10¹² - 1.5 10¹¹
Δd = 43.5 10¹¹ m
the time it takes for Neptune's light to reach Earth is
Δt = [tex]\frac{ 43.5 \ 10^{11} }{3 \ 10^8}[/tex]
Δt = 14.5 10³ s
Δt = 1.45 10⁴ s
We reduce to hours
Δt = 1.45 10⁴ s (1 h / 3600 s) = 4.03 h
Maximum away
Δd = [tex]x_{Neptune - Sum} + x_{Neptune-Sum}[/tex]
Δd = 4.50 10¹² + 1.5 10¹¹
Δd = 46.5 10¹¹
The time is
Δt = [tex]\frac{46.5 \ 10^{11}}{ 3 \ 10^8}[/tex]
Δt = 15.5 10³
Δt = 1.55 10⁴ s
We reduce to hours
Δt = 1.55 10⁴ s (1 h / 3600 s) = 4.31 h
In conclusion, the time it takes for light to arrive from Neptune to Earth varies between:
t₁ = 1.45 10⁴ s and t₂ = 1.55 10⁴ s
depending on the relative distance of the two plants
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You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
If the kinetic energy of a particle has increased to 25 times its initial value, then the percentage of the change in the wavelength which is associated with the particle's motion is...
A) 80%
B) 60%
C) 40%
D) 20%
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
How much charge is stored on two parallel-plate capacitors by the 12V battery if one is filled with air and the other is filled with a dielectric (k=3.00)
The question is incomplete, the complete question is;
how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3 m 2 and a plate separation of 2.00 mm.
The capacitance of the capacitor is the quantity of charge stored by the capacitor.
Given that;
C1= εo k * A/d
εo = permittivity of free space
C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3 m 2/2 * 10^-3
= 2.21 * 10^-11 F
C2 = 8.85 x 10-12 * 3 * 5.00×10 −3 m 2/2 * 10^-3
= 6.63 * 10^-11 F
q1 = C1V1 = 2.21 * 10^-11 C * 12 V
= 2.65 * 10^-10 C
q2 = C2V2 = 6.63 * 10^-11 F * 12 V
= 7.96 * 10^-10 C
qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C
qtotal = 1.061 * 10^-9C
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A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
A person pulls on a 9 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.5 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate
Answer:
uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no
Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
What are the systems of units? Explain each of them.
THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-
• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.
• FPS System- (Foot–Pound–Second system).
The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.
• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
What is a downside of nonprofit fitness centers?
They offer fewer luxury services than for-profit health centersThey offer fewer luxury services than for-profit health centers , ,
Very few people live near a nonprofit fitness center.Very few people live near a nonprofit fitness center. , ,
Members tend to be wealthy and lack diversity. Members tend to be wealthy and lack diversity. , ,
Membership costs more than it does in for-profit centers.Membership costs more than it does in for-profit centers. , ,
they offer fewer luxury services because they lack the money to provide them.
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.
Answer:
asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd
Explanation:
basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains
Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg
crate by a light rope. The light rope remains taut. Compared to the 6.00-kg crate,
the lighter 4.00-kg crate
Please explain why any of these multiple choices is correct!
Answer:
B. is subject to a smaller net force but same acceleration.
Explanation:
F = m*a
So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.
The net force on both crates is the same and the acceleration of both crates is the same.
The given parameters;
mass of the crate, m = 6 kgmass of the second crate, = 4 kgThe force on the 4kg crate is calculated as follows;
[tex]F_{4kg } = T + F[/tex]
The force on the 6kg crate is calculated as follows;
[tex]F_{6 kg} = -T + F[/tex]
The net force on both crates is calculated as follows;
[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]
Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.
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Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
what are the limitation of clinical thermometer
Answer:
Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
a. The molecules of a magnet are independent...
Answer:
variable
Explanation:
A bus starts from rest and accelerates at 1.5m/s squared until it reaches a velocity of 9m/s .the bus continues at this velocity and then deccelerate at -2m/s squared until it comes to stop 400m from it's starting point. how much time did the bus takes to cover the 400m?
Answer:
23s
Explanation:
s=ut+1/2at^2
the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore
400=0t+1/2(1.5)t^2
400/0.75=0.75t^2/0.75
t^2=√533.33
t=23s
I hope this helps and sorry if it's wrong
A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s
Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
On her camping trip, Penelope was in charge of collecting firewood. The firewood she found had a mass of 120 g and a volume of 480 cm3. What is the density of the firewood? Explain the steps you took to solve this problem.
Answer:
D = .25g/cm³
Explanation:
D = m/V
D = 120g/480cm³
D = .25g/cm³
A 0.2 oz. bullet leaves the muzzle of a rifle with a speed of 1420 ft/s. If the length of the barrel is 24 inches, what is the magnitude of the force acting on the bullet while it travels down the barrel
Answer:
196 lbf
Explanation:
v² = u² + 2as
1420² = 0² + 2a(24/12)
a = 1420²/4 = 504,100 ft/s²
F = ma = 0.2oz(1lb/16 oz)(1slug/32.2 lb)(504,100) = 195.6909...
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.
Answer:
Explanation:
kinematic equation (g will have a negative value if we assume UP is positive)
v² = u² + 2as
a) v = √(0² + 2(g)(y - h))
b) v = √(vi² + 2(g)(y - h))
Calculate the kinetic energy of a mass 2kg moving with a velocity of 0.1m/s
ANSWER-:
1/2 mv²
K.E = 1/2 mv²
K.E = 0.01 J.
Hence, the kinetic energy of a body is 0.01 Joule.
!! HOPE ITS HELP U !!
state any 3 properties of an ideal gas as assumed by the kinetic theory.
Answer:
The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
