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Maksim231197 [3]
3 years ago
13

An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up wit

h a velocity of +14.2 m/s and measures a time of 15.9 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet? (positive = up, negative = down).
Physics
1 answer:
bagirrra123 [75]3 years ago
6 0

Answer:

a= -1.8m/s²

Explanation:

The equation for position x for a constant acceleration a, time t, initial velocity v₀ and initial position x₀ is given by:

x=\frac{1}{2}at^2+v_0t+x_0

For the case that x = x₀, the equation becomes:

0=\frac{1}{2}at^2 +v_0t

Solving for a:

a=-\frac{2v_0}{t}

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The amount of the lighted side of the moon you can see is the same during
attashe74 [19]

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

<u>Explanation:</u>

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

5 0
3 years ago
The student throws one cannonball directly upward at 5.0 m/s and simultaneously throws the other cannonball directly downward at
Mademuasel [1]
Let the cannonball be thrown at a height of h above ground.
Then  the potential energy of the ball is
V = m*g*h
where
m = the mass of the ball
g = 9.8 m/s²

Also, the kinetic energy of the ball is
K = (1/2)mu²
where
u = 5 m/s, the vertical launch velocity.
Ignore wind resistance.

Because the total energy is preserved, the total energy (n the form of only kinetic energy) when the ball strikes the ground is
(1/2)mV²
where V = vertical velocity when the ball strikes the ground.

Expressions for both the initial and final energy are equal regardless of whether the ball s thrown downward or upward.
Therefore there is no difference in the landing speed. 

Answer: There is no difference.
8 0
4 years ago
HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Levart [38]
It is an example of "Elastic Potential Energy"

In short, Your Answer would be Option C

Hope this helps!
5 0
3 years ago
A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 7.0
krok68 [10]

Answer:

(a) T=6.07s

(b) x_{max|4.0s}=3.62cm

Explanation:

For Part (a)

The initial amplitude is given as A=7.0 cm

Apply the equation x_{max}(t)=Ae^{-t/T} with  x_{max}(7.6s)=2.0cm we have:

x_{max}(t)=Ae^{-t/T}\\2.0cm=(7.0cm)e^{-7.6s/T}\\T=-\frac{7.6s}{ln(\frac{2.0cm}{7.0cm} )} \\T=6.07s

For Part (b)

Apply x_{max}(t)=Ae^{-t/T}  with t=4.0s and T=6.07s we have

x_{max}(t)=Ae^{-t/T}\\x_{max}(4.0s)=(7.0cm)e^{-4.0/6.07}\\x_{max|4.0s}=3.62cm

8 0
4 years ago
The element lead (Pb) has a density 11.3 times that of water. Copper (Cu) has a density 7.9 times the density of water. A 5 kg m
77julia77 [94]

Answer:

The answer is B

Explanation:

Density of the element lead (Pb) is:

d_{Pb} =11,3kg/dm^3

Density of the element Copper (Cu) is:

d_{Cu} =7,9kg/dm^3

First we need o find the volume of both materials:

V_{Pb}=5/11,3=440cm^3

V_{Cu}=5/7,9=630cm^3

And the buoyant forces on elements are:

P_{Pb}=440*1*9,81/1000=4,32N

P_{Pb}=630*1*9,81/1000=6,18N

8 0
3 years ago
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