Answer:
a) Eₓ = - A y + 2B x
, b) Ey = -Ax –C
, c) Ez = 0
, d) The correct answer is 3
Explanation:
The electric field and the electric power are related
E = - dV / ds
a) Let's find the electric field on the x axis
Eₓ = - dV / dx
dV / dx = A y - B 2x
Eₓ = - A y + 2B x
b) calculate the electric field on the y-axis
Ey = - dV / dy
dV / dy = A x + C
Ey = -Ax –C
c) the electric field on the z axis
dv / dz = 0
Ez = 0
.d) at which point the electric field is zero
Since the electric field is a vector quantity all components must be zero
X axis
0 = = - A y + 2B x
y = 2B / A x
Axis y
0 = -Ax –C
.x = -C / A
We substitute this value in the previous equation
.y = 2B / A (-C / A)
.y = 2 B C / A2
The correct answer is 3
There are various ways to compute for power, and one of them is by using the voltage and current of a system. This formula is given by P = IV, where I is the current in amperes and V is voltage in volts. Using the given data above, the power in watts is calculated as follows:
P = 110(2.7) = 297 watts.
Answer:
Concepts and Principles
1- Kinetic Energy: The kinetic energy of an object is:
K=1/2*m*v^2 (1)
where m is the object's mass and v is its speed relative to the chosen coordinate system.
2- Gravitational potential energy of a system consisting of Earth and any object is:
U_g = -Gm_E*m_o/r*E-o (2)
where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.
Solution
The argument:
My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.
The counterargument:
We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):
K=1/2*m*v^2 (1)
and a gravitational potential energy Ug given by Equation (2):
Ug = -G*Mm/R
where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:
K+U=0
1/2*m*v^2 + (-G*Mm/R) = 0
1/2*m*v^2 = G*Mm/R
1/2*v^2 = G*M/R
solving for v we get
v = √2G*M/R
so we see v does not depend on the mass of the projectile
Power = (1000 kilo-Watt-hr/mo) x (1000/kilo) x (mo/30day) x (day/24 hr)
Power = (1000 x 1000 / 30 x 24) (kilo-watt-hr-mo-day/mo-kilo-day-hr)
Power = (1,000,000/720) watt
(voltage x current) = (1,000,000/720) watts
120v x current = (1,000,000/720) watts
Current = 1,000,000 / (720 x 120) Amperes
<em>Current = 11.57 Amperes</em>
Answer:
She takes pictures and analyzes DNA samples
