A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the
moving van. (2) How far is the van from its starting point? (Hint: it is the magnitude of the displacement vector).
1 answer:
Answer:
4.47 km
Explanation:
If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.
From Pythagoras theorem
AD² = AE² + ED²
Hence, the van is 4.47 km from its initial point
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Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Answer:
Average density of Sun is 1.3927 .
Given:
Radius of Sun = 7.001 × km = 7.001 ×
cm
Mass of Sun = 2 × kg = 2 ×
g
To find:
Average density of Sun = ?
Formula used:
Density of Sun =
Solution:
Density of Sun is given by,
Density of Sun =
Volume of Sun =
Volume of Sun =
Volume of Sun = 1.436 ×
Density of Sun =
Density of Sun = 1.3927
Thus, Average density of Sun is 1.3927 .
<h2>Given :</h2>
- total charge = 9.0 mC = 0.009 C
Each electron has a charge of :
For producing 1 Cuolomb charge we need :
Now, for producing 0.009 C of charge, the number of electrons required is :
_____________________________
So, Number of electrons passing through the cross section in 3.6 seconds is :
Number of electrons passing through it in 1 Second is :
Now, in 10 seconds the number of electrons passing through it is :
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