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Tju [1.3M]
4 years ago
14

A cyclist rides 6.3 km east for 21 minutes, then he turns and heads west for 6 minutes and 1.8 km. Finally, he rides east for 13

.2 km, which takes 44 minutes. (Assume motion toward the east is positive. Indicate the direction with the signs of your answers.) What is the final displacement of the cyclist?
Physics
1 answer:
Leona [35]4 years ago
6 0

Answer:

\vec{d}=17.7km

Explanation:

Displacement is a vector that defines the position of a particle. The vector extends from the initial position to the final position. Therefore, the displacement only takes into account this positions, since its trajectory is not important:

\vec{d}=6.3km-1.8km+13.2km\\\vec{d}=17.7km

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Answer: 12.98 Hz

Explanation:

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4 years ago
A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
Vika [28.1K]

Answer:

t_t=4.131\ s

Explanation:

Given:

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u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

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Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

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<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

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h'=u_y'.t'+\frac{1}{2} g.t'^2

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21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

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