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Delicious77 [7]
3 years ago
6

A 2.5 kg , 20-cm-diameter turntable rotates at 80 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turnta

ble simultaneously at opposite ends of a diameter, and stick.
What is the turntable's angular speed, in rpm, just after this event?
Physics
1 answer:
Volgvan3 years ago
3 0

Answer:

the turntable's angular speed after the event is <em>50 rpm</em>

Explanation:

Since the turntable rotates on frictionless bearings, this means that there are no external torques that will act on the table. Therefore, the table and the two blocks can be considered as an isolated system.

Therefore we will use the conservation of angular momentum which states that the initial angular momentum of the system is equal to the final angular momentum of the system.

L_i = L_f

I_i ω_i  = I_f ω_f

ω_f = (I_i ω_i) / I_f                                 (1)

Therefore, we must determine the initial moment of inertia, before dropping two blocks I_i, and the final moment of inertia, after dropping the two blocks I_f.

The moment of inertia for a disk is

I_table = 1/2 M R²

I_i = 1/2 M R²

    = 1/2 (2.0)(0.1)²

    = 0.01 kg·m²                             (2)

After dropping the blocks, we will add their moment of inertia about the axle of rotation to the moment of inertia of the table, in order to get to the final moment of inertia

I_f = ∑ m_i r_i² = I_table + I_block,1 + I_block,2

I_f = I_t + I₁ + I₂

I_f = 0.01 + (m₁ r₁²) + (m₂ r₂²)

r₁ =  r₂ and m₁ = m₂

Thus,

I_f  = 0.01 + 2 m R²

I_f = 0.01 + 2(0.5)(0.1)²

I_f = 0.02 kg·m²                              (3)

Therefore, substitute (2) and (3) into (1):

ω_f =   (I_i ω_i) / I_f      

ω_f = (0.01)(100) / 0.02

<em>ω_f = 50 rpm</em>

<em></em>

Therefore, the turntable's angular speed after the event is <em>50 rpm</em>

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A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
6. If a bicyclist, with initial velocity of zero, steadily gained speed until reaching a final velocity of 39m/s, how far would
Bezzdna [24]

Answer:

  175 m

Explanation:

The average velocity for constant acceleration is the average of the beginning and ending velocities. That is (0+39)/2=19.5 m/s. If the bicyclist rides for 9 seconds, the distance traveled is ...

  (9 s)(19.5 m/s) = 175.5 m

She would travel 175.5 meters in that time.

7 0
3 years ago
(science)<br>pls help, due today and I have alot more to do​
Effectus [21]

Answer:

I can help

Explanation:

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8 0
3 years ago
How do you determine how much kinetic energy an object has
Ierofanga [76]
I believe the answer is a. Because the formula of kinetic energy is 1/2(m)•(v^2)
3 0
2 years ago
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