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nasty-shy [4]
4 years ago
5

A boat is headed due north on a river that's flowing directly east. What is the general direction of travel for this boat with r

espect to the land?
Physics
2 answers:
nlexa [21]4 years ago
8 0

Answer:

It must be moving at an angle \theta with East towards North

Explanation:

As we know that boat headed towards North while river is flowing towards East

So here the resultant velocity of boat is vector sum of velocity of river and velocity of boat.

So we will find the vector resultant as

\vec v = \vec v_b + \vec v_r

here we know

\vec v_b = velocity of boat towards North

\vec v_r = velocity of river towards East

now if we add two vectors along North and and east then the resultant must be between North and East so the boat will move at some angle \theta with East towards North

LenKa [72]4 years ago
3 0
The answer would be northeast
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3 years ago
A 2.1 cm length of tungsten filament in a small lightbulb has a resistance of 0.067 ω. find its diameter. the resistivity of the
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The resistance of an ohmic material is given by 
R= \frac{ \rho l}{a}
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3 0
4 years ago
Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged
MArishka [77]

Answer:

A)   E = 0N/C

B)   0i + 0^^j

C)   F = 0N

D)   0^i  + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

E=k\frac{rQ}{(r+R^2)^{3/2}}               (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

F=qE=q(0N/C)=0N

D) The direction of the force is 0^i + 0^j

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iris [78.8K]

Answer:

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