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forsale [732]
3 years ago
13

6.Calculate the centripetal acceleration of the Earth towards the Sun. (r = 1.5 x 1011 m)

Physics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Given,

The radius of Sun, r=1.5\times 10^{11}\ m

To find,

The centripetal acceleration of the Earth towards the Sun.

Solution,

The centripetal acceleration of an object is given by the formula as follows :

a=\omega^2 r ....(1)

Where

\omega is angular velocity

First calculate the angular velocity.

\omega=\dfrac{2\pi}{1\ \text{day}}\\\\=\dfrac{2\pi}{24\times 3600\ s}\\\\=7.27\times 10^{-5}\ s^{-1}

Put all the values in formula (1)

a=\omega^2 r\\\\=(7.27\times 10^{-5})^2\times 1.5\times 10^{11}\\\\=792.79\ m/s^2

So, the centripetal acceleration of the Earth towards the Sun is 792.79\ m/s^2.

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Could a mixture be made only element and no compounds
Veseljchak [2.6K]

Take the tiny bit of carbon dioxide and the tiny bit of water vapor out of the air,
and the rest of what you're breathing right now is a mixture of elements.

4 0
3 years ago
Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F
Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0
4 years ago
Una cámara fotográfica analógica (no digital) tiene dos lentes intercambiables. Uno de foco 55mm y el otro de 200 mm. Toma una f
Sergeu [11.5K]

Answer:

f = 55mm,     h ’= -9.89 cm

f = 200 mm,  h ’= 42.5 cm

Explanation:

For this exercise let's start by finding the distance to the image, using the equation of the constructor

         \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distances to the object and image, respectively

lens with f₁ = 55mm = 0.55cm

         \frac{1}{q} = \frac{1}{f} - \frac{1}{p}

         \frac{1}{q_1} = \frac{1}{0.55} - \frac{1}{10}

          \frac{1}{q_1} = 1.718

          q₁ = 0.582 m

lens with f₂ = 200mm = 2m

           \frac{1}{q_2} =   \frac{1}{2} - \frac{1}{10}

            \frac{1}{q_2} = 0.4

            q₂ = 2.5 m

the magnification of a lens is given by

            m = \frac{h'}{h} = -  \frac{q}{p}

             h ’= - \frac{q}{p} \ h

let's calculate for each lens

f = 55mm

             h '= - 0.582 / 10 1.7

             h ’= 0.0989 m

             h ’= -9.89 cm

f = 200 mm

             h '= - 2.5 / 10 1.7

             h ’= -0.425 m

             h ’= 42.5 cm

The negative sign indicates that the image is real and inverted

4 0
3 years ago
..................................
sammy [17]

Hi, what is it you need help with?

8 0
2 years ago
Calculate the wavelength associated with electrons moving with p.d of 1500v
Artist 52 [7]

Answer:

3.16X10∧-11 m

Explanation:

1/2 mv2 = qV (KE = Electric potential energy)

velocity = √2qV/m = √( 2X 1.6X10∧-19 X 1500/9.11X10∧-31)

2.3X10∧7m/s

now use De Broglie equation

λ = h/mv

= 6.62X10∧-34/( 9.11X10∧-31 X 2.3X10∧7)

3.16 X 10∧-11 m

or

use the above equations and substitute to get the final eqiation

λ = h/√(2mqV) = 3.16X 10∧-11 m

4 0
3 years ago
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