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mars1129 [50]
2 years ago
15

Which solute will dissolve first in the illustration?

Physics
1 answer:
pentagon [3]2 years ago
4 0
B explanation : they are both filled to the same pint
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A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790
Nezavi [6.7K]

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

8 0
3 years ago
At an instant when the displacement is equal to a/2, what fraction of the total energy of the system is potential?
madreJ [45]
<span>At an instant when the displacement is equal to a/2, Potential energy U = 1/2ka(square) where a is displacement. when a= a/2 U = 1/4ka(square) U = E/4 Potential Energy = 1/4 Total energy</span>
4 0
3 years ago
2. If you are 5'10" tall, that is, 5 feet 10 inches, what is your height in meters? (2 54 cm = 1.00 in)
likoan [24]

Answer:D 1.7

Explanation:

Just trust me

8 0
2 years ago
Read 2 more answers
A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig
Blababa [14]

Answer:

<u>Please Mark As Brainliest!!</u>

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a)  If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf   ⇒  I₁ * ω₁  = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)  

W = 7.03 J

4 0
2 years ago
And 8 kg bowling ball is rolling along the frictionless alley
VLD [36.1K]
It will stop eventually
8 0
3 years ago
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