The figure shows the original channel of a river, as well as its current channel. Which of the following forces MOST LIKELY chan
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here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity
v = 15.4 m/s
Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
The electric field at the point exactly midway between the plates, = 0
3) The electric field, 'E', between plates is given as follows;
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, = E × e
∴ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, ≈ 1.807 × 10⁻⁷ N