Answers
Answer:
The recoil velocity of the raft is 1.205 m/s.
Explanation:
given that,
Mass of the swimmer,
Mass of the raft,
Velocity of the swimmer, v = +4.6 m/s
It is mentioned that the swimmer then runs off the raft, the total linear momentum of the swimmer/raft system is conserved. Let V is the recoil velocity of the raft.
V = -1.205 m/s
So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.
Answer:
The recoil velocity of the raft would be (pointing to the left if the swimmer runs to the right)
Explanation:
The problem states thatthe swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.
To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.
Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:
Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as
The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.
Then, from that previous relation we can clear
wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).
Related Questions
Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb
A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb
a. of water a. of bread a. of soap a. of juice a. of packet a. of sand option_piece packet grain bottle cake slice bale tank pane note tube bar can sack pile sheaf
If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don't have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart
Answers
Answer:
6.2 seconds
Explanation:
Using Newton's second law, ∑F=ma, we know the net force acting on the object is Force applied-Force of friction. The net force is 203 N. Newton's second law requires the mass of an object, not the weight force, so we will have to calculate the mass. We know that m*g=weight force, in this case, solve for the mass and you will get 210 kg. Now that we have the value of the net force and the mass, we can solve for acceleration. =0.967 m/s^2. Now, since we have the acceleration, initial velocity(0 m/s), and the final velocity (6m/s) we will use these to solve for time using the kinematic equation Vf=Vi + at. Plug in the values we know and solve for time and you will get 6.2 seconds
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?
Answers
a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.
Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:
relative to the direction of the river.
b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by
c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is
so, Dave's landing point is 540 m downstream.
d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).
Answers
Answer:
Yes, the paths of the two particles cross.
Location of path intersection = ( 1 , 2 , 3)
Explanation:
In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.
So, we can name the time of the first particle , and the time of the second particle
.
Setting the locations equal, we get the following equations to solve for and
:
Equation 1
Equation 2
Equation 3
Solving these three equations simultaneously we get:
2 seconds
4 seconds
Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.
The point of crossing can be found by using the value of or
in the location matrices. Doing this for the first particle we get:
Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )
Location of path intersection = ( 1 , 2 , 3)
Answers
Answer:
7.13559 knots
Explanation:
Maximum power = 10.5 kilowatts
where,
P = Power in kilowatts
s = Desired speed in knots
Here, P = 10.5 kW
The greatest speed of the Olympians was 7.13559 knots
Answers
Answer:
2.31 Ω
Explanation:
According to the Faraday's law of electromagnetic induction,
Induced emf = - N (dΦ/dt)
Emf = -N (ΔΦ/t)
where N = number of turns = 11
Φ = magnetic flux
ΔΦ = change in magnetic flux = 9.69 - 5.60 = 4.09 Wb
t = time taken for the change = 0.0657 s
Emf = 11(4.09/0.0657)
Emf = - 684.78 V (the minus sign indicates that the direction of the induced emf is opposite to the direction of change of magnetic flux)
From Ohm's law,
Emf = IR
R = (Emf)/I
I = current = 297 A
R = (684.78)/297
R = 2.31 Ω
Hope this Helps!!
Explanation:
Below is an attachment containing the solution.
Answers
Final answer:
To find the time rate of change of electric flux between the plates of the capacitor, use the formula \(\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}\). The displacement current between the plates can be found using the formula \(I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}\).
Explanation:
To find the time rate of change of electric flux between the plates of the capacitor, we can use the formula: \(\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}\), where \(\frac{d\phi_E}{dt}\) is the time rate of change of electric flux, \(I\) is the current, and \(A_\text{plate}\) is the area of one plate. In this case, the area of each plate is \((0.06 \,\text{m})^2\) and the current is 0.134 A. Thus, the time rate of change of electric flux is \(\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}\) V·m/s.
The displacement current between the plates of a capacitor can be found using the formula: \(I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}\), where \(I_d\) is the displacement current, \(\varepsilon_0\) is the vacuum permittivity, \(\kappa\) is the dielectric constant, and \(\frac{d\phi_E}{dt}\) is the time rate of change of electric flux. In this case, \(\varepsilon_0\) is a constant, \(\kappa\) depends on the material between the plates (not provided), and we found \(\frac{d\phi_E}{dt}\) to be \(\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}\) V·m/s. So the displacement current can be calculated once these values are known.
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Final answer:
The time rate of change of electric flux can be found using I/ε₀A. The displacement current can be found using ε₀A(dE/dt)
Explanation:
The time rate of change of electric flux between the plates of the capacitor can be found using the formula:
dΦ/dt = I/ε₀A
Where dΦ/dt is the time rate of change of electric flux, I is the current, ε₀ is the permittivity of free space, and A is the area of one of the plates.
We are given that the current is 0.134 A and the area of each plate is (0.060 m)² = 0.0036 m². Plugging these values into the equation, we get: the time rate of change of electric flux between the plates is 37.22 V·m/s.
Similarly, the displacement current between the plates can be found using the formula:
Id = ε₀A(dE/dt)
Where Id is the displacement current, ε₀ is the permittivity of free space, A is the area of one of the plates, and dE/dt is the time rate of change of electric field intensity between the plates.
We are given that ε₀ is 8.854 × 10⁻¹² F/m and dΦ E/ dt is 37.22 V·m/s. Plugging these values into the equation, we get:
Id = (8.854 × 10⁻¹² F/m)(37.22 V·m/s) = 3.29 × 10⁻¹⁰ A
Therefore, the displacement current between the plates is 3.29 × 10⁻¹⁰ A.
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