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3 years ago

14

With an average acceleration of −1.3 m/s2, how long will it take a cyclist to bring a bicycle with an initial speed of 13.0 m/s

to a complete stop?
Answer in units of s.

1 answer:

nevsk [136]3 years ago

3 0

13.0÷1.69
= seconds for the cyclist to completely stop

first solve -1.3 to the second power or -1.3 squared by multiplying -1.3 by itself
((-1.3)×(-1.3 )=1.69)
then divide 13 by 1.69
we are dividing because we are solving a deceleration equation and by multiplying 2 negatives the sum will always equal a positive and if we multiplied 13 by 1.69 we would turn this into an accelleration equation.

so divide 13 by 1.69
(13÷1.69=7.6923076923)

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Similarly we have

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3 years ago

The Doppler effect is the apparent change in frequency due to the movement of the source and the observer. This means an object

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Answer:

Doppler effect changes the wavelength of the light emitted, depending upon whether source is moving away or coming towards the observer(detector).

Explanation:

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When the source is moving away from the detector, the wavelength of the light emitted from the source appears to be increased as seen by the detector, as a result the frequency decreases(we know that frequency of light= speed of light/wavelength of light. Here speed of light is constant and frequency of light is inversely proportional to its wavelength)

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4 years ago

The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 25

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Answer:

a) $v(2\,s) = 5.9\,\frac{m}{s}$ , $v(4\,s) = -13.7\,\frac{m}{s}$ , b) $t = 2.602\,s$ , c) $h(2.602\,s) = 35.176\,m$ , d) $t = 5.281\,s$ , e) $v(2.602\,s) = -26.254\,\frac{m}{s}$

Explanation:

a) The velocity function is determined by deriving the position function in time:

$v(t) = 25.5-9.8\cdot t$

Velocities after 2 seconds and 4 seconds are, respectively:

$v(2\,s) = 5.9\,\frac{m}{s}$

$v(4\,s) = -13.7\,\frac{m}{s}$

b) The maximum height is reached when velocity is equal to zero:

$25.5-9.8\cdot t = 0$

The time when the projectile reaches the maximum height:

$t = 2.602\,s$

c) The maximum height is:

$h (2.602\,s) = 2 + 25.5\cdot (2.602\,s)-4.9\cdot (2.602\,s)^{2}$

$h(2.602\,s) = 35.176\,m$

d) The projectile hits the ground when height is equal to zero:

$-4.9\cdot t^{2}+25.5\cdot t + 2 =0$

The roots of the second order polynomial are presented below:

$t_{1} \approx 5.281\,s$

$t_{2} \approx -0.077\,s$

The first one is the only reasonable solution in physical terms.

$t = 5.281\,s$

e) The velocity of the projectile when it hits the ground is:

$v(2.602\,s) = 25.5-9.8\cdot (5.281\,s)$

$v(2.602\,s) = -26.254\,\frac{m}{s}$

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