It is symbols to make it easier and to save space!
Answer:
Our appliance and electronic energy use calculator allows you to estimate your ... using the formulas provided below; Installing a whole house energy monitoring system. ... energy consumption a few watt-hours, and you can use a monitor to estimate those too. ... Estimate the number of hours per day an appliance runs.
Explanation:
Answer:
, assuming that the gravitational field strength is
.
Explanation:
Notice that both the speed and the direction of motion of this block are constant. In other words, the velocity of this block is constant.
By Newton's Second Law, the net force on this block would be
. External forces on this block should be balanced. Thus, the magnitude of the (downward) weight of this block should be equal to the magnitude of the (upward) force that the boy applies on this block.
Let
denote the mass of this block. It is given that
. The weight of this block would be:
.
Hence, the force that the boy applies on this block would be upward with a magnitude of
.
The mechanical work that a force did is equal to the product of:
- the magnitude of the force, and
- the displacement of the object in the direction of the force.
The displacement of this block (upward by
) is in the same direction as the (upward) force that this boy had applied. Thus, the work that this boy had done would be the product of:
- the magnitude of the force that this boy exerted,
, and - the displacement of this block in the direction,
.
.
In a perfectly ELASTIC collision between two perfectly rigid objects <span>both the momentum and the kinetic energy of the system are conserved.hope it helps</span>
Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V