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1 year ago

7.5 h

Physics

1 answer:

Julli [10]1 year ago

7 0

The distance traveled by plane flying at 1200 Km/h for 2.5 hours is 3000 Km.

Velocity is a vector quantity. It has both a direction and a magnitude. Speed is used to calculate the magnitude of velocity. The meter per second is the S.I. unit for this. The units km/h and km/s are additional units. [LT-1] is the dimensional equation for it.

The distance traveled by the object is calculated as the product of the velocity with which it was moving and the time interval for which the distance covered is calculated.

Distance traveled = Velocity × Time

Given in the question

Velocity of the plane = 1200 Km/h

Time Traveled = 2.5 h

Put in the value, we get

Distance traveled = 1200 × 2.5

Distance traveled = 3000 Km

Hence, the distance traveled by plane flying at 1200 Km/h for 2.5 hours is 3000 Km.

LEARN MORE ABOUT VELOCITY HERE:

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An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

$\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

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