Answer:
Z
Explanation:
The figure represented the ray diagrams produced by a convex (converging lens). The image will appear at the point where the purple ray and the orange ray meet, so at point Z.
As we see from the picture, the image will be:
- Real (because it is behind the lens, and it can be projected on a screen)
- Inverted (the image is upside down)
- Larger than the object (we can see it by noticing that the distance between the point of intersection purple-orange ray and the axis is larger than the size of the object)
C. 4.2 mL
3.1+2.7+4.6+1.9+8.7mL=21
there are 5 numbers
21÷5=4.2
<span>Both objects receive the same impulse.</span>
Answer:
V(t) = (q0/C) * e^(−t/RC
)
Explanation:
If there were a battery in the circuit with EMF E , the equation for V(t) would be V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in V(t) (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution Vb(t)=V(t)−E . The effect of this substitution is to eliminate the E term and yield an equation for Vb(t) that is identical to the equation you solved for V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time t=0 . The solution under these conditions will look like V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time t=0 (since the capacitor was uncharged then) and rises asymptotically to E (with the result that current essentially stops flowing through the circuit).