All categories

3 years ago

At what conditions acceleration of a body is zero?

Physics

1 answer:

Marta_Voda [28]3 years ago

5 0

Answer:

when there is uniform velocity the acceleration of a body is zero

You might be interested in

The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f

tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0

3 years ago

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes t

Salsk061 [2.6K]

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{\frac{-bt}{2m}}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{\frac{-bt}{2m}}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

8 0

3 years ago

How does increasing the speed of a motorbike change the amount of energy in its kinetic energy store?

Gnesinka [82]

Answer:

Kinetic energy is energy in motion so therefor if you increase the velocity or in your case speed the kinetic energy also has to increase

Explanation:

4 0

3 years ago

Read 2 more answers

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the

Shalnov [3]

Answer:

x ’= 1,735 m, measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

x_{cm} = 1.2 -1

x_ {cm} = 0.2 m

Σ τ = 0

w₁ 1.2 + mg 0.2 - W₂ x = 0

x = $\frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}$

x = $\frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}$

let's calculate

x = $\frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}$ 2.9 1.2 + 4 0.2 / 8

x = 0.535 m

measured from the pivot point

measured from the far left is

x’= 1,2 + x

x'= 1.2 + 0.535

x ’= 1,735 m

8 0

3 years ago

si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p

forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

$E_{kin}=E_{pot}$

Donde:

$E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h$

Ahora reemplazando:

$\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} } \\\\v=5.24[m/s]$

6 0

3 years ago

At what conditions acceleration of a body is zero?​

At what conditions acceleration of a body is zero?