Answer:
Speed of water, v = 4.2 m/s
Explanation:
Given that,
Diameter of the tank, d = 17 cm
It is placed at a height of 90 cm, h = 0.9 m
We need to find the speed at which the water exits the tank through the hole. It can be calculated using the conservation of energy as :
v = 4.2 m/s
So, the speed of water at which the water exits the tank through the hole is 4.2 m/s. Hence, this is the required solution.
Answer: E= KQ/r^2
Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.
The magnitude of an electric field E, at a point is given by Coulombs law as
E/ F/q
Where F= Coulombs force exertedon the charge and q= electric charge
E= F/q=(KQq)/r^2q
E=KQ/r^2
Answer:
Explanation:
Given two vectors as follows
E₁ = 13.5 i -12 j
E₂ = -7.4 i - 4.7 j
Resultant E = E₁ + E₂
= 13.5 i -12 j -7.4 i - 4.7 j
E = 6.1 i - 16.7 j
a ) X component of resultant = 6.1 N
b ) y component of resultant = -16.7 N
Magnitude of resultant = √ ( 6.1² + 16.7² )
= 17.75 N
d ) If θ be the required angle
tanθ = 16.7 / 6.1 = 2.73
θ = 70° .
counterclockwise = 360 - 70 = 290°
Work formula:
W = F * d
F 1 = 40 N, d 1 = 6 m;
F 2 = 30 N; d 2 = 6 m.
W ( Cindy ) = 40 * 6 = 240 Nm
W ( Andy ) = 30 * 6 = 180 Nm
The difference of their amounts if work:
240 Nm - 180 Nm = 60 nm
hope it helps!
Answer:
-10.9 rad/s²
Explanation:
ω² = ω₀² + 2α(θ - θ₀)
Given:
ω = 13.5 rad/s
ω₀ = 22.0 rad/s
θ - θ₀ = 13.8 rad
(13.5)² = (22.0)² + 2α (13.8)
α = -10.9 rad/s²
