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pentagon [3]
4 years ago
9

On Venus, the acceleration due to gravity is 8.87 m / s 2 . How far would a 17 g rock fall from rest in 6.5 s if the only force

acting on it was the gravitational force due to Venus?
Physics
1 answer:
algol [13]4 years ago
7 0

Answer:

187.38 m

Explanation:

Using the equation of motion

s = ut + 1/2gt²...................... Equation 1

Where s = distance of fall, u = initial velocity of the rock, t = time taken for the rock to fall from rest, g = acceleration due to gravity of venus.

Given: u = 0 m/s ( from rest), t = 6.5 s, g = 8.87 m/s².

substituting into equation 1

s = 0(6.5) + 1/2(8.87)(6.5)²

s = 0 + 374.7575/2

s = 187.38 m.

Hence the rock will fall 187.38 m

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The star Betelgeuse is 6.1 x 10^18 m away from Earth. How old is the light we see from that star when it reaches us? There are 3
castortr0y [4]

Answer:

635 years old

Explanation:

The light reaching the earth from the sun will travel at a speed called the speed of light, and this has a universal value of 3 × 10⁸ m/s. Bearing this in mind, let us calculate the age of the light reaching the Earth from the sun:

Distance of star from Earth = 6.1 × 10⁸m

Speed of light = 3 × 10⁸ m/s

We have distance and speed, let us calculate the time of travel of the light from the star to the earth.

Distance = speed × time

6.1 × 10⁸ = 3 × 10⁸ × time

time = \frac{6.1 \times 10^{18}}{3 \times 10^8}

In order to do the division above, we will divide the whole numbers normally, then we will apply the law of indices to the power that says:

Xᵃ ÷ Xᵇ = X⁽ᵃ⁻ᵇ⁾

\therefore time = \frac{6.1 \times 10^{18}}{3 \times 10^8}\\= \frac{2.03 \times 10^{(18-8)}}{1} \\= 2.03 \times 10^{10}}\ seconds

Next, we are told that there are 3.2 × 10⁷ seconds in a year.

∴ The number of years travelled by the light from the star:

3.2\ \times 10^7\ seconds = 1\ year\\1\ second =  \frac{1}{3.2\ \times 10^7} \\\therefore 2.03 \times 10^{10}\ seconds = \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}

please note that:

2.03 × 10¹⁰ = 20300000000

3.2 × 10⁷ = 32000000

\therefore \frac{2.03 \times 10^{10}}{3.2\ \times 10^7}\\= \frac{20300000000}{32000000} \\\\= \frac{20300}{32} \\= 634.347\ years\\

The closest answer in the option is 635 years, and we are short of this by some points due probably to approximations in the calculation.

8 0
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Answer:

B is correct. Electrons are added to the rod.

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