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g100num [7]
3 years ago
15

A boat is travelling down stream in a river at 20 m/s south. A passenger on the boat walks toward the back of the boat at 10 m/s

north. Relative to the ground, at what velocity is the passenger moving?
A) 10 m/s north
B) 10 m/s south
C) 30 m/s north
D) 30 m/s south
Physics
1 answer:
artcher [175]3 years ago
8 0

The question is unclear regarding the boat's velocity. Is it 20 m/s south relative to the water, or relative to the earth? (It is a river, after all...)

There's also the possibility that the boat's velocity relative to the river is 0. Take the south direction to be negative and north to be positive, and denote by v_{A/B} the velocity of a body A relative to a body B. Under these conditions,

v_{B/E}=v_{B/W}+v_{W/E}\iff-20\,\dfrac{\mathrm m}{\mathrm s}=0+-20\,\dfrac{\mathrm m}{\mathrm s}

(B for boat, E for earth, W for water) so that the passenger's velocity relative to the earth is

v_{P/E}=v_{P/B}+v_{B/W}+v_{W/E}

(P for passenger)

v_{P/E}=10\,\dfrac{\mathrm m}{\mathrm s}+0-20\,\dfrac{\mathrm m}{\mathrm s}=-10\,\dfrac{\mathrm m}{\mathrm s}

or 10 m/s south.

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Answer:

Below

Explanation:

First draw the vectors that represent both electric fields.

E1 is the elictric field created by q1, E2 is the one created by q2.

● q1 is negative so E1 will point from P.

● q2 is positive so E2 will point out of P

(Picture below)

■■■■■■■■■■■■■■■■■■■■■■■■■■

The resulting electric field is equal to the sum of the two fields since both vectors are colinear.

Let E be the total field.

● E = E1 + E2

The formula of the electric field intensity is:

● E = K ×(q/d^2)

-K is Coulomb's constant

-d is the distance between the charge and the object ( here P)

-q is the charge

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E1 = K × (q1/d1^2)

The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)

Coulombs constant is 9×10^9 m^2/C^2

● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]

● E1 = -359.43 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E2 = K ×(q2/d^2)

The distance between q2 and P is 0.25 m.

● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]

● E2 = 463.68 N/C

■■■■■■■■■■■■■■■■■■■■■■■■■■

● E = E1 + E2

● E = -359.43+463.68

● E = 105.25 N/C

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