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stellarik [79]
3 years ago
13

(a) A small object with mass 3.75 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 2.55 m centered

at the origin. It starts at the point with position vector 2.55 i m. Then it undergoes an angular displacement of 8.95 rad.
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?
Physics
1 answer:
Eddi Din [679]3 years ago
8 0

Answer:3.95 m/s

Explanation:

Given

mass of object m=3.75 kg

\omega =1.55 rad/s

radius of circle =2.55 m

initial Position r=2.55 \hat{i}

angular displacement \theta _0=8.95 rad

8.95 radian can be written as

1.42 (2\pi )

i.e. Particle is at first quadrant with \theta =0.4242\pi \times \frac{180}{\pi }

\theta =76.36^{\circ}

(c)velocity is v=\omgea \times r

v=1.55\times 2.55=3.95 m/s

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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
anzhelika [568]

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

5 0
2 years ago
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