Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
Therefore, the height of the cliff is 121.276 m
Answer:
<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>Explanation:
Given :
Current A
Time sec
We know that the current is the rate of flow of charge.
From the formula of current,
<h3>Where charge transfer between cloud and earth.
C
Hence, the charge transferred from the cloud to earth is 1 Coulomb.
The electric field at one corner of a square is 1614217 N/C.
Explanation:
The distance between x and y direction diagonals.
As per the given details the distance between diagonals is calculated as
0.5² + 0.5² = c² => c = 0.707 m
Charge to the right: In x direction
In order to find the electric charge towards x direction
we use e = kq/r² formula
As 'k' is coulomb's constant it's value is 9 x N m²/C²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.707)²] cos 45
e = 225000√2 N/C
X direction sum = 1218198 N/C.
Similarly as shown in x direction the charge is same for y direction also
Charge below: For y direction
e = kq/r²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.5)²] sin 45
e = 159099 N/C
Y direction sum = 1059099 N/C
Resultant electric field strength:
1218198 ² + 1059099² = e²
e = 1614217 N/C [45 degrees below the horizontal]