Answer:
L = 1.545 m
Explanation:
Let the total length of the rod is L
now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum
so we will have

so we have

F = 663 N


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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.
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In this question ,
surface vertical force = Weight of the object
Thus ;
svf = ( mass ) × ( gravity acceleration )
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If gravity acceleration is 10 :
svf = 10 × 10 = 100 N
So ;
frictional force = 100 × 0.20
frictional force = 20 N
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If gravity acceleration is 9.8 :
svf = 10 × 9.8 = 98 N
So ;
frictional force = 98 × 0.20
frictional force = 19.6 N
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D because I learned this 2 years ago
Answer : The value of the constant for a second order reaction is, 
Explanation :
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = ?
t = time = 17s
= final concentration = 0.0981 M
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:


Therefore, the value of the constant for a second order reaction is, 
Answer:
0.00016 kg
Explanation:
Given:
Power = P = 1.2 × 10⁹ Watts
Power = work done / Time
efficiency = 0.30
Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W
Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules
E = m c² , where c is the speed of light and m is the mass.
⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²
= 0.00016 kg