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sashaice [31]
4 years ago
10

during a car ride Sandra travel led directly east for 20 miles and then directly south for 15 miles what was her displacement. 2

5 miles at 59.0 degrees south of east, 25 miles at 25.9 degrees east of south, 25 miles at 36.9 degrees south of east, or 25 miles at 42.5 degrees east of south?
Physics
1 answer:
damaskus [11]4 years ago
8 0
25 miles at 42.5 degrees east of south. (20^2+15^2 =C^2...You get 25 miles...East of South because she traveled a longer distance to the East. Then you find the angle between the original and ending point.) Hope it helps. You may ask me any other questions.
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A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal
Gemiola [76]

Answer:

Time taken, T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

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T=\dfrac{mg}{cos\theta}

Sum of forces in x direction,

T\ sin\theta=\dfrac{mv^2}{r}

mg\ tan\theta=\dfrac{mv^2}{r}.............(1)

Also, r=l\ sin\theta

Equation (1) becomes :

mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}

v=\sqrt{gl\ tan\theta.sin\theta}...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

T=\dfrac{2\pi r}{v}

Put the value of T from equation (2) to the above expression:

T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}

T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}

On solving above equation :

T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}

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4 0
3 years ago
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Answer:

- Decreasing the resistance

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Explanation:

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Thus, to increase the conductance, we have to decrease the resistance.

Resistance here has a formula of;

R = ρL/A

Where;

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Thus, to decrease the resistance, we will have to use a shorter length and smaller area of wire.

8 0
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