Answer: 1.5×10^10 N/C
Explanation:
E= F/q
Where E= magnitude of the electric field
F= force of attraction
q= charge of the given body
Given F= 6.5×10^-8 N
q= 4.3× 10^-18 C
Therefore, E = 6.5×10 ^-8/ 4.3×10^-18
E = 1.5×10^10 N/C
Answer: It is not likely.
Explanation:
When the bus is moving forward, all the objects inside of it also are moving forward.
Now, as the objects inside the buss are not fixed to the bus, if the bus suddenly stops the objects inside of it will keep moving forward, because of the conservation of the momentum, defined as the quantity of motion (Similar to when you are in a car and it suddenly stops, you can feel the forward impulse).
Then is not likely that, in a case where the bus stops suddenly, an object inside the bus flies backward in opposite direction to the previous movement of the bus.
<span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
Answer:
2.9 M
Explanation:
The concentration-time equation for a second order reaction is:
1/[A] = kt + 1/[A°]
Where,
A = concentration remaining at time, t
A° = initial concentration
k = rate constant
1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81
1/[A] = 0.345
= 1/0.345
= 2.9 M.
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))