Answer:
149 m
Explanation:
The distances across the lake is forming a triangle.
let the distance between the point and the left side be 'x'
and the distance between the point and the right be 'y'
and the distance across the lake be 'z' and the angle opposite to 'z' be 'Z' given:
∠Z = 83°
x = 105 m
y = 119 m
Now, applying the Law of Cosines, we get
z² = x² + y² - 2xycos(Z)
Substituting the values in the above equation, we get
z² = 105² + 119² - 2×105×119×cos(83°)
or
z = √22140.48
or
z = 148.796 m ≈ 149 m
The point is 149 m across the lake
The centripetal acceleration is 
Explanation:
For an object in uniform circular motion, the centripetal acceleration is given by

where
v is the speed of the object
r is the radius of the circle
The speed of the object is equal to the ratio between the length of the circumference (
) and the period of revolution (T), so it can be rewritten as

Therefore we can rewrite the acceleration as

For the particle in this problem,
r = 2.06 cm = 0.0206 m
While it makes 4 revolutions each second, so the period is

Substituting into the equation, we find the acceleration:

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Answer:
16 cm
Explanation:
Given that,
The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.
Let the x-axis to be the +ve and on the right side and -ve on the left
Thus, displacement would be:
= 0 -3 + 7 -6
= -2 cm
This implies that the object displaces 2cm towards the left.
While the total distance covered by the object equal to,
= 0cm + 3cm + 7cm + 6cm
= 16 cm
Thus, <u>16 cm</u> is the total distance.
Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
To solve the two parts of this problem, we will begin by considering the expressions given for gravitational potential energy and finally kinetic energy (to find velocity). From the potential energy we will obtain its derivative that is equivalent to the Force of gravitational attraction. We will start considering that all the points on the ring are same distance:

Then the potential energy is

PART A) The force is excepted to be along x-axis.
Therefore we take a derivative of U with respect to x.



This expression is the resultant magnitude of the Force F.
PART B) The magnitude of loss in potential energy as the particle falls to the center

According to conservation of energy,

