Let F = the downstream speed of the water.
<span>Then the boat's upstream speed is: 15 - F </span>
<span>The boat's downstream speed is: 15 + F </span>
<span>Assume both the journeys mentioned take T hours, then using "speed x time = distance" we get: </span>
<span>Downstream journey: (15 + F)T = 140 </span>
<span>Upstream journey: (15 - F)T = 35 </span>
<span>Add the two formulae together: </span>
<span>(15 + F)T + (15 - F)T = 140 + 35 </span>
<span>15T + FT + 15T - FT = 175 </span>
<span>30T = 175 </span>
<span>T = 35/6 </span>
<span>Use one of the equations to find F: </span>
<span>(15 + F)T = 140 </span>
<span>15 + F = 140/T </span>
<span>F = 140/T - 15 </span>
<span>F = 140/(35/6) - 15 </span>
<span>F = 24 - 15 </span>
<span>F = 9 </span>
<span>i.e. the downstream speed of the water is 9 kph </span>
<span>Therefore, the boat's speed downstream is 15 + F = 15 + 9 = 24 kph.
the answer is: *24kph*</span>
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The potential energy of the object depends on
- the height of the object with respect to some reference points,
- the mass of the object,
- the gravitational field the object is in.
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Hope it helps ~
Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:

B = 7.24 * 10⁻⁶T
B = 7.24 μT
For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:
F = ILB
F = magnetic force, I = current, L = wire length, B = magnetic field strength
Given values:
F = 0.60N, L = 1.0m, B = 0.20T
Plug in and solve for I:
0.60 = I(1.0)(0.20)
I = 3.0A