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Katyanochek1 [597]
3 years ago
11

Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time

it takes to paddle a canoe across the river, as long as the boat is pointed perpendicular to the bank CofOthe river.
Physics
1 answer:
anzhelika [568]3 years ago
5 0

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let \overset \rightarrow {v_y} represent the velocity of the boat across the river in the direction, \overset \rightarrow {d_y}, and let, \overset \rightarrow {v_x}, represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = \overset \rightarrow {v_y}

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is \overset \rightarrow {v_y}, such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

\overset \rightarrow {v_y} = 3 m/s

\overset \rightarrow {v_x} = 4 m/s

The width of the river, w_y = 6 meters, we have;

  • The resultant velocity = \sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s

The direction, θ, is given as follows;

\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}

The length of the path of the boat, <em>l</em>, is given as follows;

l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = \dfrac{l}{v}, therefore;

  • t = \dfrac{10}{5} = 2
  • The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

t = \dfrac{w_y}{v_y}

Therefore;

t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

brainly.com/question/15907242

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If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.
Schach [20]

Answer:

The magnitude of each charge is 2.82\times10^{-6}\ C

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

F=\dfrac{kq^2}{r^2}

Where, q = charge

r = separation

Put the value into the formula

20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}

q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}

q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}

q=2.82\times10^{-6}\ C

Hence, The magnitude of each charge is 2.82\times10^{-6}\ C

6 0
3 years ago
Find the ratio of effusion rates of hydrogen gas and krypton gas.
algol13

Answer:

RE of Hydrogen = 6.47 x RE of Krypton

Explanation:

Actually the correct formula for comparing rate of effusion (RE) of two gases is:

RE of Gas A

------------------- = √ ( Molar mass of B / Molar mass of A)

RE of Gas B

You can designate which of the two gases you have (hydrogen and krypton) will be your gas A and gas B. So for this particular problem, let us make hydrogen as gas A and Krypton as gas B. So the equation becomes:

RE of Hydrogen

------------------------- = √ (Molar mass of Krypton / Molar mass of Hydrogen)

RE of Krypton

Get the molar masses of Hydrogen and Krypton in the periodi table:

RE of Hydrogen

------------------------- = √ (83.798 g/mol / 2 g/mol)

RE of Krypton

RE of Hydrogen

------------------------- = 6.47  ====> this can also be written as:

RE of Krypton

RE of Hydrogen = 6.47 x RE of Krypton

It means that the rate of effusion of Hydrogen gas will be 6.47 faster than the rate of effusion of Krypton gas. With the type of question you have, it doesn't matter which gases goes on your numerator and denominator. What's important is that you show the rate of effusion of a gas with respect to the other. But if that's concerns you the most, then take the gas which was stated first as your gas A and the latter as your gas B unless the problem tells you which one will be on top and which is in the bottom.

3 0
3 years ago
A 45 kg object has a momentum of 225 kg-m/s northward. What is the object's velocity?
melomori [17]

Answer: 5 m/s North

Explanation:

Velocity = Momentum/Mass

Velocity = 225 kg*m/s/45kg

Velocity = 5 m/s North

3 0
2 years ago
The water waves below are traveling along the surface of the ocean at a speed of 2.5 m/s and
gavmur [86]

Answer:2 seconds

Time= Distance/speed

=50/25

=2seconds

3 0
3 years ago
Jeff is listening to the radio. He knows that the radio produces sound energy because
Likurg_2 [28]

Answer:

A.  he can hear music coming from the radio.

Explanation:

Jeff knows that the radio produces sound energy because he can hear music coming from the radio.

The music from the radio is a form of sound energy.

  • Sound energy is a form of energy dissipated from a body and can be discerned by humans.
  • Sound is a mechanical wave that requires a material medium for its propagation.
  • It travels by series of rarefaction and compression along its path of motion.
  • Since sound energy is an energy in motion, it is a kinetic energy.
  • It travels parallel to its source and therefore, it is a longitudinal wave.
8 0
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