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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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The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the
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Answer:

27.5 days

0.92 month

Explanation:

r = radius of the orbit of moon around the earth = 3.85\times10^{8} m

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According to Kepler's third law, Time period is related to radius of orbit as

T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}

inserting the values, we get

T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec

we know that

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6 0
3 years ago
Hai vận động viên chạy trên cùng 1 đoạn đường, vận động
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Answer:

Pemain A

Explanation:

Mengingat data berikut;

Kecepatan pemain A = 12 m/s

Kecepatan pemain B = 36 km/h

Untuk menentukan siapa pelari tercepat di antara dua pemain;

Pertama-tama, kita harus mengubah kecepatan menjadi satuan standar pengukuran yang sama.

Jadi, mari kita gunakan pengukuran umum meter per detik.

Konversi:

36 km/h = (36 * 1000)/(60 * 60)

36 km/h = 36000/3600

36 km/h = 10 m/s

Kecepatan pemain B = 10 m/s

Oleh karena itu, dibandingkan dengan kecepatan pemain A; pemain A lebih cepat.

6 0
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B, a disurbance would make it unstable and affect the center of mass, which would then affect the equilibrium.

4 0
3 years ago
Calculate the force of gravity on the same 1 kg mass if it were 6.4x10^6 m above earth's surface
vova2212 [387]
The force of gravity on a certain object is calculated through the equation,
 
   F = Gm1m2 / r²

where F is the force, G is a constant, m1 and m2 are masses of the object and Earth, respectively and r is the distance. Substituting the known values for this item,

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Answer: 9.37 N
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