Mr. Hitch taught us about sedimentary, metamorphic, and igneous rocks. He described how they were formed, what they contain, and showed us samples of each. He is a good geologist.
The missing word and answer is: geologist.
Answer:
X-Positions: Y-Positions
x(0) = 0 y(0) = 0
x(2) = 120 m y(2) = 19.6 m
x(4) = 240 m y(4) = 78.4 m
x(6) = 360 m y(6) = 176.4 m
x(8) = 480 m y(8) = 313 m
x(10) = 600m y (10) = 490 m
Explanation:
X-Positions
- First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
- After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
- So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:
- It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.
Y-Positions
- We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
- As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
- At any moment, it is subject to the acceleration of gravity, g.
- As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:
- Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
- y(2) = 2* 9.8 m/s2 = 19.6 m
- y(4) = 8* 9.8 m/s2 = 78.4 m
- y(6) = 18*9.8 m/s2 = 176.4 m
- y(8) = 32*9.8 m/s2 = 313.6 m
- y(10)= 50 * 9.8 m/s2 = 490.0 m
Answer:
oa
Explanation:
it may be oa is the right answer for this question
but I don't know properly
Answer: 3.63 km/s
Explanation:
The escape velocity equation for a craft launched from the Earth surface is:
Where:
is the escape velocity
is the Universal Gravitational constant
is the mass of the Earth
is the Earth's radius
However, in this situation the craft would be launched at a height over the Eart's surface with a space elevator. Hence, we have to add this height to the equation:
Finally:
<span>gravitational force = 0.1 mg-wt. = 0.1 * 10^-6 * 9.8 N = Gm1m2/r^2
m1 = 40 kg m2 =15 kg and r = 0.2 m
Put in and find G</span>