Answer:
The magnetic force on the section of wire is 
.
Explanation:
Given that,
Current 
Length = 0.750 m
Magnetic field 
We need to calculate the magnetic force on the section of wire
Using formula of magnetic force
Since, 
Hence, The magnetic force on the section of wire is .
Answer: Option B.
Since here the truck is moving on a circular track, it will experience centripetal force.
F(centripetal) = m × acc
or
where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton
To find = radius of track=?
r = 81.94 m
Therefore, radius of track is 81.94 m
Answer:
1 ohm
Explanation:
First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:
1 / Eq = (1 / r₁) + (1 / r₂)
The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:
Eq = r₁ + r₂
Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:
1/E₁ = 1/2 + 1/2
1/E₁ = 1
E₁ = 1Ω
1/E₂ = 1/2 + 1/2
1/E₂ = 1
E₂ = 1Ω
This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:
E₃ = E₁ + E₂ = 1 + 1 = 2Ω
The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:
1/Eq = 1/E₃ + 1/2
1/Eq = 1/2 + 1/2
1/Eq = 1
Eq = 1Ω
I see the light moving exactly at speed equal to c.
In fact, the second postulate of special relativity states that:
"The speed of light in free space has the same value c<span> in all inertial frames of reference."
</span>
The problem says that I am moving at speed 2/3 c, so my motion is a uniform motion (constant speed). This means I am in an inertial frame of reference, so the speed of light in this frame must be equal to c.
A. 9 J
In a force-distance graph, the work done is equal to the area under the curve in the graph.
In this case, we need to extrapolate the value of the force when the distance is x=30 cm. We can easily do that by noticing that there is a direct proportionality between the force and the distance:
where k is the slope of the line. We can find k, for instance chosing the point at x=5 cm and F=10 N:
And now we can calculate the work by calculating the area under the curve until x=30 cm, F=60 N:
B. 24.5 m/s
The mass of the arrow is m=30 g=0.03 kg. The kinetic energy of the arrow when it is released is equal to the work done by pulling back the bow for 30 cm:
where m is the mass of the arrow and v is its speed. By re-arranging the formula and using W=9 J, we find the speed:
