Answer:
Here's what I get
Explanation:
A. Distance between A and B.
h = -½gt²
The stones go faster the farther they fall.
Stone A has already reached 5 m when B is released.
When B reaches 5 m, A has dropped further and is falling even faster.
The distance between the stones increases with time.
Figure 1 shows this effect in a graph of height vs. time.
B. Speed of Stone B
v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²
v = 9.9 m/s
The stone is travelling at 9.9 m/s when it reaches 5 m.
C. Velocity vs time
v = -gt
Both stones accelerate at the same rate.
When Stone B has reached 10 m at time t, Stone A is falling much faster.
Fig. 2 shows this in a graph of velocity vs time.
Answer:
option b
Explanation:
There is an object pulled inward in an electric field.
We have to find out of the four options given which is true.
a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.
b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object
c) The object has a negative charge will be correct only if the original charge was positive hence wrong
d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge
So only option b is right
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
maximumforce is F = mg
Explanation:
For this case we must use Newton's second law,
Σ F = m a
bold indicate vectors, so we will write it in its components x and y
X axis
Fₓ = maₓ
Axis y
Fy - W = m a
Now let's examine our case, with indicate that the bird is level, the force of the wings can have a measured angle with respect to the x axis, where the vertical component is responsible for the lift, let's use trigonometry to find the components
Cos θ = Fₓ / F
Fₓ = F cos θ
sin θ = Fy / F
Fy = F sin θ
Let's replace and calculate
F sin θ -w = m a
As the bird indicates that leveling at the same height, so the vertical acceleration is zero (ay = 0)
F sin θ = w = mg
The maximum value of this equation occurs when the sin=1, in this case
F = mg
Complete Question:
Check the circuit in the file attached to this solution
Answer:
Total current = 0.056 A(From left to right)
Explanation:
Let the current in loop 1 be I₁ and the current in loop 2 be I₂
Applying KVL to loop 1
30 - (I₁ - I₂)500 + I₂R + 15 = 0
45 - 500I₁ - 500I₂ + RI₂ = 0
I₁ = 30mA = 0.03 A
45 - 500(0.03) - 500I₂ + RI₂ = 0
30 -500I₂ + RI₂ = 0...............(1)
Applying kvl to loop 2
-RI₂ - 15 + 10 - 400I₁ = 0
-RI₂ = 5 + 400*0.03
RI₂ = -17 ................(2)
Put equation (2) into (1)
30 -500I₂ -17 = 0
-500I₂ = 13
I₂ = -13/500
I₂ = -0.026 A
The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026
Total current = 0.056 A
The current will flow from left to right
