The force require to keep grouper submerged is 8.207N.
According to Archimedes principle buoyant force of any object must equal to weight of fluid it displaced.
The expression for the force exerted to stay submerged in salt water is
F = F(b) - w(fish)
where F(b) = buoyant force
w(fish) = weight
now substitute w(b) for F(b)
→ F = Vρg - w(fish)
where V = volume of sea water
ρ = density of sea water
Now by Archimedes principle V = m(fish) / ρ(fish)
→ F = (m(fish) / ρ (fish) ) ρg - m(fish)g
F = (85 kg/1015 kg-m^-3) (1.025× 10³ kg-m^-3) (9.8 m/s^2)
- (85kg) × 9.8 m/s^2
F = 841.207N - 833N
F = 8.207 N
Hence, the force require to keep grouper submerged is 8.207N.
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Answer:
C)the right cord pulls on the pulley with greater force than the left cord
Explanation:
As we can see the figure that B is connected to the right string while A is connected to the left string
Now force equation for B as it will move downwards is given as
similarly for block A which will move upwards we can write the equation as
now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks
so here pulley will rotate clockwise direction
So tension in the right string must be more than the left string
So correct answer will be
C)the right cord pulls on the pulley with greater force than the left cord
The elapsed time when the particle returns to the origin is determined from the ratio of initial velocity and acceleration of the particle.
<h3>Time of motion of the particle</h3>
The time of motion of the particle is calculated by applying Newton's second law of motion.
F = ma
F = m(v)/t
where;
- t is time of motion of the particle
- m is mass of the particle
- v is velocity of the particle
a = v - u/t
v = u + at
when the particle returns to the origin, direction of u, = negative.
final velocity = 0
0 = -u + at
at = u
t = u/a
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