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jekas [21]
3 years ago
13

a 2 meter tall astronaut standing on mars drops her glasses from her nose. how long will the astronaut have before he hits the g

round?
Physics
1 answer:
lorasvet [3.4K]3 years ago
7 0
Here,
Height (S) = 2m
Gravity on mars (g) = 3.7m/s^2
Initial velocity (u) = 0 m/s^2
By the one of the formula of the motion,
S = ut + 1/2at^2
2 = 0 * t + 1/2*3.7*t^2
2 = 1.85t^2
t^2 = 2/1.85 = 1.081
t =1.03s
So, it will take 1.03s long..

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Aperture is measured in F-stops, in which the f-stops is the amount of light allowed to pass through the aperture, which simply put means that the smaller the aperture, the higher the f-stops. What it does is reduce the amount of light that reaches the film, so the higher the f-stops, the less light reaches the film.
4 0
3 years ago
Determine the electrical force of attraction between two balloons with separate charges of +3.5 × 10-8 C and -2.9 × 10-8 C when
kenny6666 [7]

Answer:

The electrical force of attraction between the balloons is 2.16\times 10^{-5}\ N.

Explanation:

Given that,

Charge 1, q_1=+3.5\times 10^{-8}\ C

Charge 2, q_2=-2.9\times 10^{-8}\ C

Distance between the charges, d = 0.65 m

We need to find the electrical force of attraction between two balloons. It is given by the formula as :

F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{3.5\times 10^{-8}\times 2.9\times 10^{-8}}{(0.65)^2}\\\\F=2.16\times 10^{-5}\ N

So, the electrical force of attraction between the balloons is 2.16\times 10^{-5}\ N. Hence, this is the required solution.

3 0
3 years ago
The weakest type of friction that occurs between solid surfaces is _____ friction
Evgen [1.6K]

Answer:

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Explanation:

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Hopefully this helped.
5 0
4 years ago
A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
blondinia [14]

Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

8 0
4 years ago
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