Answer: 5.30m
Explanation:
depth of pool = 3.2 m
i = 67.75°
Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
n₁ = 1, n₂ =1.33, r= 44.09°
Hence,
Distance of Google from edge if pool is:
2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) =5.30m
Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>
Answer:
(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.
(b) 
Explanation:
Given:
- mass of the child,
- height of the slide,
- initial velocity of the child at the slide,
- final velocity of the child at the bottom of slide,
(a)
∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.
Initial potential energy:
Kinetic energy at the bottom of the slide:
(b)
Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.
This can be given as:
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
