Answer:
5
Explanation:
Let there are three capacitances, C1, C2, and C3.
Combination I:
All the three are connected in series combination.
Combination II:
All the three are connected in parallel combination.
Combination III:
C1, C2 are in parallel and then C3 in series.
Combination IV:
C1, C3 are in parallel and then C2 in series.
Combination V:
C3, C2 are in parallel and then C1 in series.
This is a defective, misleading question, and should never be asked in a Physics class.
There is no such thing as the force due to the impact.
If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.
It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.
I would kick this question back to the Physics teacher. It's meaningless,
and the longer you try to work on it, the more nonsense you'll plant into
your head that'll need to be dug out later.
Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.
m = mass of the truck = 23 00 kg
v = speed of the truck down the highway = 32 m/s
K = kinetic energy of the truck = ?
kinetic energy of the truck down the highway is given as
K = (0.5) m v²
inserting the values
K = (0.5) (2300) (32)²
K = (0.5) (2300) (1024)
K = (1150) (1024)
K = 1177600 J
hence the kinetic energy of the truck comes out to be 1177600 J
It can help with measurements and when you want to add measurements to a cylinder or a beaker so ya
