Answer:
see I need help for chemistry pls anyone here helps me I must submit before 10:30
Answer:
(1) - (2)
and
(3) - (4)
Explanation:
Hi!
The Newton's third law states that for every action there's and equal and opposite reaction.
Lets us consider the first force:
<em>(1) the force of the horse pulling on the cart</em>
Therefore, its counterpart, will be a force of the cart pulling on the horse, which is actually:
<em> (2) the force of the cart pulling on the horse</em>
<em />
Analogously, the third force:
<em>(3) the force of the horse pushing on the road</em>
and the fourth force:
<em>(4) the force of the road pushing on the horse </em>
<em />
From a action reaction pair, since the former force acts on the road by the horse, and the latter on the horse by the road
It's kinda tough, since we don't know the actual numerical locations of the points, only an approximate picture.
The big ball has 11 times the mass of the small ball, so the small ball is 11 times as far from the barycenter as the big ball is.
If any of the points is marked at the actual barycenter, it can only be point-A .
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
Answer:
1. 21.66 Ohms
2. 3.38 A
3. 6.7 V
Explanation:
1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)
Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)
Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms
2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)
3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)
I hope this helps. I am not an expert in physics but its ok :)
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