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MariettaO [177]
3 years ago
15

Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the

ir masses are different. show answer No Attempt How are they able to balance on the seesaw? They are able to balance their individual energies. All of these. Their densities are the same. They are able to balance forces due to gravity. They are able to balance their individual momenta. They are able to balance torques due to gravity.
Physics
1 answer:
Westkost [7]3 years ago
5 0

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

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3 years ago
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe
Bogdan [553]

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

4 0
3 years ago
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

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The German sociologist Max Weber formulated a three-component theory of stratification that defines a status group (additionally status class and status estate) as a group of humans within a society who can be differentiated by means of non-financial characteristics along with honor, prestige, ethnicity, race, and religion.

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