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Alex Ar [27]
2 years ago
14

Jim can ride his bike at 12 meters per second. Bob can ride his bike at 10.5 m/s. If they

Physics
1 answer:
Vinvika [58]2 years ago
3 0

Answer:

just a example

Explanation:

Elena always rides her bicycle at a speed of 15 miles per hour. On Sunday, she goes on a 24 mile bike ride. How many hours does this ride take?

Solution

The speed of 15 miles per hour is a rate. The key words that tell us that this is a rate are "speed" and "miles per hour". We can write:

r = 15

Next, 24 miles is a distance. We have:

d = 24

Now use the d=rt equation to get

24 = 15t

To solve this, divide both sides by 15 to get

t = 24/15

Both are divisible by 3, so this fraction reduces to

t = 8/5 = 1.6

Elena's ride takes 1.6 hours.

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A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
  • initial displacement position of equilibrium, u(0)=0

<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

∵ at t=0;\ u'(0)=20\ cm.s^{-1}\ \Rightarrow B=\frac{1}{5}

u(t)=\frac{1}{5} sin\ (25t)

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