At the bottom of the tank :
P = ρgH
P = (1000 kg/m³)(10 m/s²)(1 m)
P = 10000 N/m²
F = P • A
F = (10000 N/m²)(1 m²)
F = 10000 N
At the side of the tank :
Pav = ½ρgH
Pav = ½(1000 kg/m³)(10 m/s²)(1 m)
Pav = 5000 N/m²
F = P • A
F = (5000 N/m²)(1 m²)
F = 5000 N
Answer: 1- A diagram showing the forces acting on the object.
2- The force vector describes a specific amount of force and its direction. You need both value and direction to have a vector.
3- A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.
4- I don’t know
5- When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces
6- By applying an unbalanced force, you can change the motion of an object. Unbalanced forces can make an object at rest start moving, make a moving object stop, or change the direction and speed of the object.
7- If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force.
8- Imbalance or unequal force
9- The result is no motion. Balanced forces can cancel each other out. Any time there is a balanced force, the object does not move.
10- net force is the vector sum of all the forces that act upon an object.
RESTATE THESE
Explanation:
RESTATE!!!!
Answer:
Explanation:
Given a square side loop of length 10cm
L=10cm=0.1m
Then, Area=L²
Area=0.1²
Area=0.01m²
Given that, frequency=60Hz
And magnetic field B=0.8T
a. Flux Φ
Flux is given as
Φ=BA Sin(wt)
w=2πf
Φ=BA Sin(2πft)
Φ=0.8×0.01 Sin(2×π×60t)
Φ=0.008Sin(120πt) Weber
b. EMF in loop
Emf is given as
EMF= -N dΦ/dt
Where N is number of turns
Φ=0.008Sin(120πt)
dΦ/dt= 0.008×120Cos(120πt)
dΦ/dt= 0.96Cos(120πt)
Emf=-NdΦ/dt
Emf=-0.96NCos(120πt). Volts
c. Current induced for a resistance of 1ohms
From ohms law, V=iR
Therefore, Emf=iR
i=EMF/R
i=-0.96NCos(120πt) / 1
i=-0.96NCos(120πt) Ampere
d. Power delivered to the loop
Power is given as
P=IV
P=-0.96NCos(120πt)•-0.96NCos(120πt)
P=0.92N²Cos²(120πt) Watt
e. Torque
Torque is given as
τ=iL²B
τ=-0.96NCos(120πt)•0.1²×0.8
τ=-0.00768NCos(120πt) Nm
