Beryllium has a charge of 2, and bromine has a charge of –1. which is the best name for the ionic bond that forms between them?
1 answer:
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Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g = 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f =
⇒ T = 4M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × ×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
Answer:
Explanation:
When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by , according to the equation
where is the specific heat capacity of the substance.
In this problem, we have:
is the amount of heat supplied to the sample of gold
m = 0.1 kg = 100 g is the mass of the sample
is the specific heat capacity of gold
Solving for , we find the change in temperature
And since the final temperature was
The initial temperature was