Answer:
c. Smaller than.
Explanation:
The energy stored in a capacitor is given as
E = 1/2CV²...................... Equation 1
Where E = Energy stored in a capacitor, C = capacitance of the capacitor, V = Voltage.
Also,
C = εA/d ......................Equation 2
Where ε = permitivity of the material, A = cross-sectional area of the plate, d = distance of separation of the plate.
substitute equation 2 into equation 1
E = 1/2εAV²/d ................. Equation 3
From equation 3 above,
The energy stored in a capacitor is inversely proportional to the distance of separation between the plates.
Hence when the plate is pulled apart by a distance D (D>d) The energy stored in the capacitor will be smaller.
The right option is c. Smaller than.
Answer:
6.9m/s
Explanation:
Given parameters:
Acceleration of the object = 4m/s²
Distance = from x; 2m to x; 8m
Unknown:
Average velocity = ?
Solution:
From the given parameters, we use the right motion equation to solve the problem.
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance covered
Distance = 8m - 2m = 6m
Initial velocity = 0m/s
The final velocity gives us the average velocity in this problem;
v² = 0² + (2 x 4 x 6) = 48
v = √48 = 6.9m/s
The correct answer is the amoeba will deploy its pseudopods (cytoplasmic extentions) to capture the prey and phagocyte.
The amoeba most known and probably the most representative of the kind. Large (up to 500 microns), common in stagnant waters, extremely voracious as evidenced by multiple digestive vacuoles.
Amoebae are characterized by a deformable cell body emitting changes of shape, the pseudopods, which allow them to crawl on a support or to capture microscopic prey by phagocytosis.
Answer:
d = 10.076 m
Explanation:
We need to obtain the velocity of the ball in the y direction
Vy = 24.5m/s * sin(35) = 14.053 m/s
To obtain the distance, we use the formula
vf^2 = v0^2 -2*g*d
but vf = 0
d = -vo^2/2g
d = (14.053)^2/2*(9.8) = 10.076 m
Answer:
<em>c. The astronaut does not need to worry: the charge will remain on the outside surface.</em>
<em></em>
Explanation:
The astronaut need not worry because <em>according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside.</em> This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.
