Answer: The 6 kg rock sitting on a 3.2 m cliff.
Explanation:
The potential energy of an object of mass M that is at a height H above the ground us:
U = M*H*g
where g is the gravitational acceleration:
g = 9.8m/s^2
Then:
"An 8 kg rock sitting on a 2.2 m cliff"
M = 8kg
H = 2.2m
U = 8kg*2.2m*9.8 m/s^2 = 172.48 J
"a 6 kg rock sitting on a 3.2 m cliff"
M = 6kg
H = 3.2m
U = 6kg*3.2m*9.8m/s^2 = 188.16 J
You can see that the 6kg rock on a 3.2m cliff has a larger potential energy.
Answer:
mass of the other asteroid = ![4.49*10^9kg\\](https://tex.z-dn.net/?f=4.49%2A10%5E9kg%5C%5C)
Explanation:
We use the definition for the force between two celestial objects under the action of the gravity they produce using newton's general gravitational constant: ![G=6.674*10^{-11} \frac{N*m^2}{kg^2}](https://tex.z-dn.net/?f=G%3D6.674%2A10%5E%7B-11%7D%20%5Cfrac%7BN%2Am%5E2%7D%7Bkg%5E2%7D)
The force between the two asteroids will then be given by:
![F_G=G*\frac{M_1*M_2}{d^2}](https://tex.z-dn.net/?f=F_G%3DG%2A%5Cfrac%7BM_1%2AM_2%7D%7Bd%5E2%7D)
where G is Newton's gravitational constant, the asterioid's masses are M1 and M2 respectively, and d is the distance between them.
We replace the known values in he equation above, and solve for the missing mass:
![F_G=G*\frac{M_1*M_2}{d^2}\\1.05*10^{-4}N=6.674*10^{-11} \frac{N*m^2}{kg^2} \frac{3.5*10^6kg*M_2}{(10^5m)^2} \\1.05*10^{-4}=2.3359*10^{-14} * M_2\\M_2=\frac{1.05*10^{-4}}{2.3359*10^{-14}} =4.49*10^9kg](https://tex.z-dn.net/?f=F_G%3DG%2A%5Cfrac%7BM_1%2AM_2%7D%7Bd%5E2%7D%5C%5C1.05%2A10%5E%7B-4%7DN%3D6.674%2A10%5E%7B-11%7D%20%5Cfrac%7BN%2Am%5E2%7D%7Bkg%5E2%7D%20%5Cfrac%7B3.5%2A10%5E6kg%2AM_2%7D%7B%2810%5E5m%29%5E2%7D%20%5C%5C1.05%2A10%5E%7B-4%7D%3D2.3359%2A10%5E%7B-14%7D%20%2A%20M_2%5C%5CM_2%3D%5Cfrac%7B1.05%2A10%5E%7B-4%7D%7D%7B2.3359%2A10%5E%7B-14%7D%7D%20%3D4.49%2A10%5E9kg)
Since the units for the given quantities are all in the SI system, our resultant units for the unknown mass of the asteroid will be in kg.
The cardboard would be able to act as a better cell wall since it can act more closely to an actual cell wall. A cell wall has layers and covers the plant cells in order to protect it. The cardboard can cover it more with its layers rather than the jump rope.
Answer: µ=0.205
Explanation:
The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,
f=Fw
The sum of the moments about the base of the ladder Is 0
ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²
Note that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.
0= 0.9637FwL - (67.91L)2.652
Fw=180.1/0.9637
Fw=186.87N
f=186.81N
Since Fw=f
We know Fw, so we know f.
But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (25.8 + 67.08)kg * 9.8m/s² =
910.22N
so
µ = f / Fn
186.81/910.22
µ= 0.205
Answer:
The energy released is 5.31 10¹⁴ J
Explanation:
Einstein found an expression to relate the changes of energy matter that is
ΔE = Δm C²
This energy is called energize rest, in case of matter in motion the term of kinetic energy must be added
We can use this equation to find the transformation of the matter in energy
ΔE = 5.9 10⁻³ (3 10⁸)²
ΔE = 5.31 10¹⁴ (Kg m² /s²)
ΔE = 5.31 10¹⁴ J
The energy released is 5.31 10¹⁴ J