Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0
Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
Based on the situation above the the work done was 400 Joules. <span>Q = FS cos(theta) is the so-called work function. It's important to learn the work physics; you'll see it over and over in science/physics class. Theta is the angle between the force vector F and the distance vector S. In your problem we assume theta = 0, the two vectors were assumed aligned.</span>
Answer:
time=4s
Explanation:
we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion
, where max=
Given that, at i(2)=
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
Now substitute
⇒
⇒
⇒
Applying logarithm on both sides,
⇒
⇒
⇒
now subs.
⇒
also
⇒
⇒
To calculate the mass of the fuel, we use the formula
Here, m is the mass of fuel, V is the volume of the fuel and its value is and is the density and its value of 0.821 g/mL.
Substituting these values in above relation, we get
Thus, the mass of the fuel 247 .94 kg.
Answer:
<h2>
0.147136N/m²</h2>
Explanation:
Pressure is defines as force exerts by a body per its unit area.
Pressure = Force/Area
Given the total force exerted by the gas = 6.05N
Area of the rectangular container = 0.121 m * 0.201 m = 0.024321m²
Pressure of the sample = 6.05/0.02432
Pressure of the sample = 0.147136N/m²