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Alex73 [517]
4 years ago
8

The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the

object moves from small R to large R (like a rocket being launched) or from large R to small R (like a comet approaching the earth).
O True O False
Physics
1 answer:
ValentinkaMS [17]4 years ago
5 0

Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R  to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

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A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti
allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q =  −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0
3 years ago
Please Help with this
Tresset [83]

Answer: c is correct

Explanation: i did this

3 0
3 years ago
Help Meeee Please!!!!! This was due so long ago... DO NOT REPORT, THERE IS NO REASON TO REPORT! Assignment: Weathering in Action
zaharov [31]

Answer: Chemical

Explanation: because it is changeing its chemical structure if its wrong sorry .

3 0
3 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
Please help this is really important thanks
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