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hoa [83]
3 years ago
15

I need help on this.​

Physics
1 answer:
lana [24]3 years ago
3 0

Answer:

The speed change during the 45-minute trip is 20[mph]

Explanation:

When we see the speed at the 45 minutes this is 20 [mph] and at the 0 minutes the speed is 0 [mph].

Therefore the change is (20 - 0) = 20 [mph]

In the attached image we can see the different figures. In fig 1 we can see the bicycle's speed after 10 minutes when the speed becames constant.

In the fig. 2 we can find the graph when the biker stopped at 30 minutes and took a 15-minute break.

Figures 3 and 4, show the differences when a horizontal line is traced on a position vs time graph, and when the horizontal line is traced in a speed vs time graph.

For fig 3 we can conclude that the body is not moving therefore there is no velocity or acceleration. And for the fig 4, we can realize that the area under the horizontal line represents a displacement during the respective interval of time.

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A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 43.0 vibrations
blsea [12.9K]

Answer:

0.218

Explanation:

Given that

Total vibrations completed by the wave is 43 vibrations

Time taken to complete the vibrations is 33 seconds

Length of the wave is 424 cm = 4.24 m

to solve this problem, we first find the frequency.

Frequency, F = 43 / 33 hz

Frequency, F = 1.3 hz

Also, we find the wave velocity. Which is gotten using the relation,

Wave velocity = 4.24 / 15

Wave velocity = 0.283 m/s

Now, to get our answer, we use the formula.

Frequency * Wavelength = Wave Velocity

Wavelength = Wave Velocity / Frequency

Wavelength = 0.283 / 1.3

Wavelength = 0.218

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How long will it take you to ride your bike 120 miles if you are riding at 5mph?
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Read 2 more answers
A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball
adell [148]

Answer:

Part a)

T = 2\sqrt{\frac{R}{3g}}

Part b)

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

v_y = \sqrt{Rg/3}

Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

\theta_i = 33.7 degree

Part f)

H = \frac{13R}{8}

Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

initial speed = v_i

angle = \theta_i degree

Now the two components of the velocity

v_x = v_i cos\theta_i

similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}

maximum height is given as

H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}

so we have

v_i sin\theta = \sqrt{Rg/3}

time of flight is given as

T = \frac{2v_isin\theta_i}{g}

T = \frac{2\sqrt{Rg/3}}{g}

T = 2\sqrt{\frac{R}{3g}}

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}

\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

so maximum range is

X = \frac{v^2}{g}

X = \frac{13R}{4}

3 0
3 years ago
Help me bro, thanks all​
bagirrra123 [75]
C beautiful yellow silk
D was sleeping
3 0
3 years ago
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