Ask question

Ask question

All categories

3 years ago

11

Transparent

1 answer:

marusya05 [52]3 years ago

4 0

Answer:

A..................

You might be interested in

The cylinder of gravity of cylinder is where

Elodia [21]

Explanation:

In uniform gravity it is the same as the centre of mass. For regular shaped bodies it lies at the centre of the that particular body. Hence for a cylinder centre of gravity lies at the midpoint of the axis of the cylinder.

5 0

4 years ago

Read 2 more answers

Which method of measurement would be accurate but lack precision?

kodGreya [7K]

Answer:

B

Explanation:

reading the volume of water in a graduated cylinder which can be read to the nearest mL is accurate, it lacks precision due to the bottom meniscus formed.

the bottom meniscus may cause a wrong reading due to refraction of light

7 0

2 years ago

A grapefruit falls from a tree and hits the ground 0.72 s later.

xxTIMURxx [149]

Answer:

<em>The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s</em>

Explanation:

<u>Free Fall Motion </u>

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.8 m/s^2.$

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

$\displaystyle y=\frac{gt^2}{2}$

Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:

$\displaystyle y=\frac{9.8\cdot 0.72^2}{2}$

y = 2.54 m

The final speed is computed below:

$vf=9.8\cdot 0.72$

vf = 7.06 m/s

The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s

6 0

3 years ago

A friend on skis stands still on level ground, with dry 0.00°C with a coefficient of static friction of 0.0300. How hard would y

Salsk061 [2.6K]

Answer:

20.6 N

Explanation:

Friction equals normal force times coefficient of friction.

F = Fn μ

On level ground, normal force equal weight.

Fn = W

Therefore:

F = W μ

F = (685 N) (0.0300)

F = 20.6 N

6 0

3 years ago

A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l

son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

$\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Using formula of focal length for glass ethyl alcohol system

$\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Divided equation (II) by (I)

$\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}$

Where, $n_{g}$ = refractive index of glass

$n_{a}$ = refractive index of air

$n_{ethyl}$ = refractive index of ethyl

Put the value into the formula

$\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}$

$\dfrac{f'}{11.5}=\dfrac{25}{7}$

$f'=\dfrac{25}{7}\times11.5$

$f'=41.07\ cm$

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

7 0

4 years ago