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mrs_skeptik [129]
3 years ago
15

A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles

east for one hour. What is the instantaneous velocity at any time during the second hour of the trip?
35 mph east
42 mph east
60 mph east
85 mph east
Physics
2 answers:
egoroff_w [7]3 years ago
6 0

60 mph east...........

Ivan3 years ago
4 0

Answer:

The instantaneous velocity at any time during the second hour of the trip is 60 mph east

(C) is correct option

Explanation:

Given that,

Distance in first hour = 25 miles

Distance in second on hour = 60 miles

Instantaneous velocity :

Instantaneous velocity is equal to the distance divided by the time

We need to calculate the instantaneous velocity at any time during the second hour of the trip

Using formula of instantaneous velocity

.v = \dfrac{d}{t}

Where, d = distance

t = time

Put the value into the formula

v = \dfrac{60}{1}

v= 60\ miles/hr

The direction of the instantaneous velocity is in east .

Hence, The instantaneous velocity at any time during the second hour of the trip is 60 mph east

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At some instant and location the electric field associated with an electromagnetic wave in vacuum has the strength 96.5 V/m. Fin
prisoha [69]

Answer:

B = 32.17 x 10^-8 Tesla

u = 8.24 x 10^-8 J/m^3

P/A = 24.72 W/m^2

Explanation:

E = 96.5 V/m

velocity of light, c = 3 x 10^8 m/s

Let B be the magnetic field.

The relation between the electric field strength and the magnetic field strength is given by

B = E / c = 96.5 / (3 x 10^8) = 32.17 x 10^-8 Tesla

Let u be the energy density.

u = \frac{1}{2}\times \varepsilon _{0}E^{2}+\frac{1}{2\mu _{0}}B^{2}

u = \frac{1}{2}\times 8.854\times 10^{-12}\times 96.5\times 96.5+\frac{1}{2\times 4\times 3.14\times 10^{-7}}\times 32.17^{2}\times 10^{-16}

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Let Power flow per unit area is

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7 0
3 years ago
A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome
Leya [2.2K]

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

L=I\omega

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Put the value into the formula

L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320

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(B). Rotation momentum of sphere is same rotational momentum of the  flywheel

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Using formula of rotational momentum

L_{sphere}=L_{flywheel}

I\omega_{sphere}=I\omega_{flywheel}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}

Put the value into the formula

\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}

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Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

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