Answer:
Zero work done,since the body isn't acting against or by gravity.
Explanation:
Gravitational force is usually considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.
Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.
The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.
Magnetism is <span>a physical phenomenon produced by the motion of electric charge, resulting in attractive and repulsive forces between objects.</span>
Answer:
B) The cosmic background radiation is expected to contain spectral lines of hydrogen and helium, and it does.
Explanation:
<h2>Answer with Explanation </h2>
Dalton’s theory can be classified by the following hypotheses:
1) All material was formed of particles, unbreakable and strong construction segments.
2) All particles of a given component are indistinguishable in volume and characteristics
3) Compounds are determined by a mixture of two or more distinct kinds of atoms.
4) Chemical responses appeared in the rearrangement of the reacting atoms.
This theory was to explain all matter in terms of atoms and their characteristics, the law of conservation of volume and the law of constant composition.
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 =
/ W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s