A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used?

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

half of a circle

explanation:

the arrow and the top of the circle is matching together. it is not a polygon because a polygon would always be the same shape. therefore, an half of a circle is called semicircle so you would find the radius of it.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards