There are 3 bases before you reach home plate.
The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
<u />
Learn more about equilibrium of forces here:
brainly.com/question/6995192
Answer: C.
Explanation:
For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.
Now, the stored energy can be written as:
E = (1/2)*Q^2/C
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.
Answer:
Gonads
Explanation:
The gonads, the primary reproductive organs, are the testes in the male and the ovaries in the female. These organs are responsible for producing the sperm and ova, but they also secrete hormones and are considered to be endocrine glands.
Hope this helps :)
<u>Given data</u>
Source temperature (T₁) = 177°C = 177+273 = 450 K
Sink temperature (T₂) = 27°C = 27+273 = 300 K
Energy input (Q₁) = 3600 J ,
Work done = ?
We know that, efficiency (η) = Net work done ÷ Heat supplied
η = W ÷ Q₁
W = η × Q₁
First determine the efficiency ( η ) = ?
Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)
= 33.3% = 0.333
Now, Work done is W = η × Q₁
= 0.33 × 3600
<em> W = 1188 J</em>
<em>Work done by the engine is 1188 J</em>
