Answer:
The answer is D) diagonal line with varying slope, from 3 to 5 on USATestprep
Explanation:
Insert moves the insertion point to the beginning of data in a cell so the answer is INSERT :)))
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Answer:
'A ball is thrown straight up with an initial speed of 12 m/s_ What are the velocity and acceleration when it is at the top of its trajectory? Select all apply. v=12 mls a = 0 v =-12 mls a = 9.8 m/s2 Oa=-9.8 m/s2'
Explanation:
I look it up
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Answer:
Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.
Explanation:
i want the answer i don't know
Answer:
Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e 
= 875 kg.m²
We know from the conservation of angular momentum that:
![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
From the conservation of energy as well;we have :
^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)
