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3 years ago

What is the angular velocity of a 75.0-cm-diameter tire on an automobile traveling at 90.0 km/h?

Physics

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

Angular velocity of tire $\approx 66.67\ s^{-1}$

Explanation:

Given:

Diameter of tire of an automobile =75.0 cm

Linear velocity of automobile = 90.0 km/hr

To find angular velocity of tire.

Diameter of tire in meters = $75\ cm \times \frac{1\ m}{100\ cm}= 0.75\ m$

Radius of tire = $\frac{1}{2}\times diameter=\frac{1}{2}\times 0.75\ m = 0.375\ m$

Linear velocity in meters per second = $\frac{90\ km}{1\ hr}\times \frac{1000\ m}{1\ km}\times \frac{1\ hr}{3600\ s}= 25\ m/s$

Angular velocity $\omega$ is given by :

$\omega=\frac{v}{r}$

where $v$ represents linear velocity and $r$ represents radius of tire.

Plugging in values.

$\omega=\frac{25\ m/s}{0.375\ m}$

∴ $\omega=66.666 s^{-1}\approx 66.67\ s^{-1}$ (Answer)

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when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how m

scoray [572]

Answer:

Time, t = 3.2 ms

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4 years ago

Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac

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Explanation:

Relation between electric field and charge density is as follows.

E = $\frac{\sigma}{2 \epsilon}$

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So, $E_{\text{outside}}$ = 0

$E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}$

or, $E_{inside} = \frac{\sigma}{\epsilon}$

Now, formula to calculate the potential difference of two conductors is as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

It is given that,

d = 6.0 mm = $6 \times 10^{-3} m$

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Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

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3 years ago