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3 years ago

6

What will be the weight of a box, mass 10Kg on the surface of a hypothetical planet having mass double of Earth and the radius i

s 3 times of Earth

1 answer:

vlada-n [284]3 years ago

3 0

Answer:

W ’= 21.78 kg

Explanation:

The expression for weight is

W = m g

let's look for the acceleration of gravity with the universal law of gravitation

F = G m M / r2

F = m (G M / r2)

without comparing the two equations

g’= G M / r2

in that case M = 2 Mo and r = 3 ro

where mo and ro are the mass and radius of the earth

we substitute

g ’= G 2Mo / (3r₀) 2

G ’= 2/9 G Mo / r₀²

g ’= 2/9 g

the weight of the body on this planet is

W ’= m g’

W ’= m 2/9 g

let's calculate

W ’= 2/9 10 9.8

W ’= 21.78 kg

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goldenfox [79]

We are given the acceleration of the bus as a function of time:

a(t) = 1.2t

Let the velocity also be a function of time v(t).

Since a(t) is the change of v(t) over time, we can use the fundamental theorem of calculus to determine the velocity at t = 2s, or v(2), given that:

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v(1) = 5

v(2) - v(1) = $\int\limits^2_1{a(t)}\,dt$

v(2) - 5 = $\int\limits^2_1{1.2t}\,dt$

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3 years ago

Which describes a velocity?

SpyIntel [72]

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Explanation:

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2 years ago

Read 2 more answers

A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pi

anzhelika [568]

Answer:

The intensity of light be maximum is for angles 23.3° and 52.3°.

Explanation:

Given that,

Wave length = 632.8 nm

Distance = 1.60 μm

We need to calculate the intensity of light be maximum

Using Bragg's law

$d\sin\theta=n\lambda$

$\theta=\sin^{-1}(\dfrac{n\lambda}{d})$

We need to calculate the angle for different value of n

Using Bragg's law

$\theta=\sin^{-1}(\dfrac{n\lambda}{d})$

For n₁,

Put the value into the formula

$\theta=\sin^{-1}(\dfrac{632.8\times10^{-9}}{1.60\times10^{-6}})$

$\theta=23.3^{\circ}$

For n₂,

$\theta=\sin^{-1}(\dfrac{2\times632.8\times10^{-9}}{1.60\times10^{-6}})$

$\theta=52.3^{\circ}$

Hence, The intensity of light be maximum is for angles 23.3° and 52.3°.

6 0

3 years ago

An amplifier produces a sound wave with a power of 1.00 μW (micro-Watt). If it is a point source and radiates uniformly in all d

gtnhenbr [62]

Answer:

(a) $I=8.84*10^{-9}W/m^{2}$

(b) $\beta =39.46dB$

Explanation:

Given data

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The point is 3m away from the point source

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The intensity at a certain distance from a point source that emits sound wave is given as:

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For Part (b)

The Sound level is given by

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Where

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Substitute the given values to find Sound level

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4 0

3 years ago

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klio [65]

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We can solve this problem using the <u>Poiseuille equation</u>:

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$\eta=3(10)^{-3} Pa.s$ is the viscosity of blood

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4 years ago