Answer:

Explanation:
It is given that,
Distance between the spotlight and the wall, y = 24 m
Height of the woman, h = 2 m
The woman walks toward the building at the rate of 0.6 m/s, 
In the attached figure, triangle ABC and MNC are similar. So,
............(1)

When she is 2 meters from the building. So x = 24-2 = 22 m

Differentiating equation (1) i.e.




So, her shadow is decreasing at the rate of 0.059 m/s. Hence, this is the required solution.
The final magnification will be 400-fold or 400 times the original size of the object.
For magnifying smaller objects, a compound microscope is used.
A compound microscope consists of an objective and an eyepiece, whose diagram is shown in the adjoining image.
The lens near the object is called an objective and the other one is the eyepiece.
Let the magnification of the objective be m1
Let the magnification of the eyepiece be m2
The final magnification by the microscope, M, will be
M = m1 x m2
Putting the values in the above equation
M = 40 x 10
M= 400
Thus, the final magnification will be 400-fold or 400 times the original size of the object.
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The large truck can rest at a stable equilibrium as compare to the small,from the definition of inertia, <span>Inertia is the tendency of an object to remain at rest or in motion.</span>
Answer:
9 hours
Explanation:
For each battery:
Current = 1 A
Voltage = 1.5 V
current and EMF required = 5.0 A and 4.5 V
Since each battery has a current of 1 A, we'll need to link five of them in parallel with our load to get a current of 5 A. so that the total current equals 5 amps
. Now that each battery has a 1.5 V voltage drop and we require an emf of 4.5 V, we would use three batteries in series rather than a single battery. So, there is a need to must substitute every single battery with three batteries in series for all five single batteries linked in parallel.
So, the total no. of batteries is
= 5 × 3 = 15 batteries.
On each battery, the charge is = 3 amp hours
∴
The total charge = 15 × 3 = 45 amp hours.
Since the charge transferred within 1 hour = 5 amp
Then, the required lifetime of the battery is:
= 45 amp-hours/ 5 amp
= 9 hours
Answer:
11.625 Ohm
Explanation:
Let V be the Voltage charge of the loop, as this is constant we know that before the resistor addition the current I is:
V/R1 = 1.9 or V = 1.9R1
After the resistor addition to series R = R1 + 3.1
I = V/R = V/(R1 + 3.1) = 1.5
We can substitute V = 1.9R1
1.9R1 = 1.5R1 + 1.5*3.1
0.4R1 = 4.65
R1 = 4.65/0.4 = 11.625 Ohm