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3 years ago

14

Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad

ius is 6.37 x 10^6 m.

1 answer:

Xelga [282]3 years ago

6 0

0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is 1.4129 x 10⁻⁸ the radius of the Earth.

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What types of materials can make up sediments?

GREYUIT [131]

Answer: Pieces of minerals, rocks, plant and animal remains.

Explanation: Pieces of minerals, rocks, plant and animal remains. whether or not patterns cause flow rates of rivers to vary. sand settles from faster-moving water;smaller costs of silt and clay that form up mud settle from slower moving water.

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3 years ago

If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

mixas84 [53]

Answer: $3.874 m/s^2$

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

$60mi/h \approx 26.8224m/s$

$34mi/h \approx 15.1994 m/s$

we know acceleration is given by $=\frac{velocity}{Time}$

$a=\frac{15.1994-26.8224}{3}$

$a=-3.874 m/s^2$

negative indicates that it is stopping the car

Distance traveled

$v^2-u^2=2as$

$\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s$

$s=\frac{488.419}{2\times 3.874}$

s=63.038 m

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3 years ago

In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

cluponka [151]

Answer:

The thickness is $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 480 \ nm = 480*10^{-9} \ m$

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The second order of dark fringe considered is $m_2 = 6$

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$y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=> $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=> $\Delta y = 2.4 *10^{-6} \ m$

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3 years ago

An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the cu

Alex

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

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R = V/I

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R = 10 ohms

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R_eq = 10 + 5

R_eq = 15 ohms

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