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bekas [8.4K]
3 years ago
7

An insect sit's on the edge of a spinning record that has

Physics
1 answer:
attashe74 [19]3 years ago
5 0

Answer:

1.67rad/s^2

Explanation:

a= v^2/r

a = 0.5^2/0.15

a= 1.67 rad/s^2

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Can anybody tell me the right answer ? please and thank you !!
ad-work [718]

Answer:

c. selenium

Explanation:

sulfur and selenium are in the same group

3 0
3 years ago
After clearing the bar on a high jump, you land softly on a giant mattress. landing on a mattress is more comfortable than landi
madam [21]

Answer:

The right approach is Option b (the force..................exert on you).

Explanation:

  • Even before you fall on something like a soft object, users eventually slow to a halt. You are still giving away all the downward momentum, but progressively although with small powers, you are doing so.  
  • Although you can get injured by massive powers, this gradual displacement is a positive thing. And that is why you have a mattress you would like to settle on.  

The other options given are not connected to the situation described. So, the solution here was the right one.

3 0
3 years ago
A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is
Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_{f} = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_{f}

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_{f} = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

5 0
3 years ago
A 94.7 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.75 r
Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

Step 1:  Given data

Mass of the horizontal circular platform = 94.7 kg

Mass of the monkey = 21.1 kg

Initial angular velocity = 1.75 rad/s

A monkey drops a 9.25 kg bunch of bananas

They hit the platform at 4/5 of its radius from the center

Model the platform as a disk of radius 1.63 m

Step 2: Calculate the moment of inertia of the disk

I = ½ * m * r² = ½ * 94.7 * 1.63² = 125.80

Step 3: Calculate the initial angular momentum

I = 125.80 * 1.75 = 220.15

Step 4: Calculate the moment of inertia for the bananas

For the bananas, r = 4/5 * 1.63 = 1.304 m

I = 9.25 * 1.304² = 15.73

Step 5: Calculate Moment of inertia for the monkey

I = 21.1 * 1.63² = 56.06

Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

3 0
3 years ago
A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
satela [25.4K]

Nah a a yuloure

Explanation:

5 0
3 years ago
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