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Annette [7]
4 years ago
13

An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 0.4 cm/s at an angle of 52◦ with the lon

g edge. How long will it take her to get across? Answer in units of s.
Physics
1 answer:
nordsb [41]4 years ago
5 0

Answer:

t=108.50s

Explanation:

The ant is under an uniform motion. So, we get the time taken by the ant to cross the strip from the definition of speed:

v=\frac{x}{t}\\t=\frac{x}{v}

In this case, we use the vertical speed of the ant (v_y=vsin\theta), since the ant crosses the strip that is 34.2 cm wide (vertically). So:

t=\frac{y}{v_y}\\t=\frac{y}{vsin\theta}\\t=\frac{34.2cm}{0.4\frac{cm}{s}sin(52^\circ)}\\t=108.50s

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The spillage of water

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What is the focal length of concave mirror that magnifies , by a factor of +3.2 , an object that is placed 30cm from the mirror?
guapka [62]

Answer:

23 cm

Explanation:

The formula for magnification is;

Magnification = image distance / object distance

use values as;

3.2 = v / 30   where v is image distance

v =30*3.2

v=96 cm

The relationship of the focal length with image distance and object distance is expressed as;

\frac{1}{u} +\frac{1}{v} =\frac{1}{f}

where f is the focal length and u is object distance

use values in the equation as;

\frac{1}{30} +\frac{1}{96 }  =\frac{1}{f}

\frac{1}{f} =\frac{7}{160}

f=160/7

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7 0
3 years ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

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3 years ago
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