Answer:
The thrown rock will strike the ground 
earlier than the dropped rock.
Explanation:
<u>Known Data</u>
, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use 
, to find the total time of fall, so 
, then clearing for 
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use , to find the total time of fall, so 
, then, 
, as it is a second-grade polynomial, we find that its positive root is 
Finally, we can find how much earlier does the thrown rock strike the ground, so 
Answer:
one-third of its weight on Earth's surface
Explanation:
Weight of an object is = W = m*g
Gravity on Earth = g₁ = 9.8 m/s
Gravity on Mars = g₂ = 
g₁
Weight of probe on earth = w₁ = m * g₁
Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )
As g₂ = g₁/3 --------- ( 2 )
Put equation (2) in equation (1)
so
Weight of probe on Mars = w₂ = m * g₁ /3
Weight of probe on Mars = m * g₁ = w₁
⇒Weight of probe on Mars = Weight of probe on earth
Answer:
B. Mechanical energy= 50J+30J=80J
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Answer:
B. Outside the nucleus.
Explanation:
Electrons orbit the nucleus of the atom.
