To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.
Fg = m • g
Fg = 1.3 kg • 9.8 m/s^2
Fg = 12.74 N or about 12.7 N.
The solution is C. 12.7 N.
closed and unsecure circuit
Option c) 1.5 V
Explanation:
<em>As the circuit is build in series first we will find the current passing through the complete circuit. Current stays the same in each element is the series cirucuit, however, the voltage is different.</em>
Voltage is given by the following formula:
V = IR
<em>Because we have to find current through whole circuit, we will first find resistance of the whole circuit.</em>
Equivalent Resistance R(eq): R1 + R2 = 60 + 60 = 120 ohm
Current passing through whole circuit be:
= 0.025
Now we will find out the voltage between C and D:
Current stays the same in series circuit: I = 0.025 c
Resistance between C and D is, R = 60 ohm
Voltage becomes, V = IR = 0.025 * 60 = 1.5 V
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
