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iren [92.7K]
3 years ago
10

If we always drop the balls from 1-m

Physics
1 answer:
dusya [7]3 years ago
3 0

Answer:

height

weight of ball

time of ball falling

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Nonconjugated β,γ-unsaturated ketones are in base-catalyzed equilibrium with their conjugated α,β-unsaturated isomers. The mecha
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Answer:

Explanation:

The equilibrium mechanism for the reversible acid is catalyzed by the isomerization of non conjugated β, γ- unsaturated ketones, like 3-cyclohexanone to their conjugated α, I²- unsaturated isomers.

Oxygen of the Carbonyl group in the ketone is protonated by the acid and this is followed by the abstraction of an α- hydrogen from the protonated 3-cyclo hexanone to yield ethanol

2-cyclo hexanone can be obtained by acid catalyzation of 3-cyclohexanone isomers through the formation of it's "enol".

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A block slides down a frictionless plane having an inclination of 15.0°. The block starts from rest at the top, and the length o
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3 years ago
How do geologists construct the geologic column?​
Luda [366]

Answer:

They do this by joining pieces of different Rock sequences together from different places in the whole world.

Explanation:

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3 years ago
The raising and lifting the ball by kicking is called
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The raising and lifting the ball by kicking is a B) Scoop

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The wheel of a car has a radius of 20.0 cm. It initially rotates at 120 rpm. In the next minute it makes 90.0 revolutions. (a) W
just olya [345]

Answer:

Explanation:

Given that,

Radius of the wheel, r = 20 cm = 0.2 m

Initial speed of the wheel, \omega_i=120\ rpm=753.98\ rad/s

Displacement, \theta=90\ rev=565.48\ rad

To find,

The angular acceleration and the distance covered by the car.

Solution,

Let \alpha is the angular acceleration of the car. Using equation of rotational kinematics as :

\theta=\omega_i t+\dfrac{1}{2}\alpha t^2

565.48=753.98\times 60+\dfrac{1}{2}\alpha (60)^2

\alpha =-24.81\ rad/s^2

Let t is the time taken by the car before coming to rest.

t=\dfrac{\omega_f-\omega_i}{\alpha }

t=\dfrac{0-753.98}{-24.81}

t = 30.39 seconds

Let v is the linear velocity of the car. So,

v=r\times \omega_i

v=0.2\times 753.98

v = 150.79 m/s

Let d is the distance covered by the car. It can be calculated as :

d=v\times t

d=150.79\ m/s\times 30.39\ s

d = 4582.5 meters

or

d = 4.58 km

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