Which group has the Same electron shells? A Titanium and Gold B Selenium and Tin C Calcium and Magnesium D Lithium and Boron

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

masya89 [10]

The velocity would be the same for all ramps.

5 0

3 years ago

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Reading or writing on a phone is especially dangerous while driving because:

algol [13]

Answer:

Explanation:

When we are driving we need a lot of attention and concentration. Also one involved in driving should be consious and courteous

Thus, whenever a person is drives, and when he is disactracted by Mobile phones it will destroy his presence of mind.

It will good if use mobile after stopping the vehicle

Thanks

5 0

3 years ago

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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

3 years ago

_____ is the horizontal change between two points on the line.

sattari [20]

The horizontal change between two points on a graph is called the 'run'.

The vertical change between two points is called the 'rise'.

6 0

3 years ago

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Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619

=

t

+

C

Here C =constant of integration.

x

=

0 at t

=

0

, we get: C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

t

−

4.7619

.

(

2

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

=

1

−

4.7619

= −

3.7619

or x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in m

i

l

/

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

v

=

7.5×

Now differentiating above equation w.r.t time we get:

a

=

d

v/

d

t

=

−

0.675

/

At

t

=

0

the joggers acceleration is :

a

=

−

0.675

m

i

l

/

=

−

4.34

×

f

t

/

(c) required time for the jogger to run 6 miles is obtained by setting

x

=

6 in equation (2). We get:

−

4.7619

(

1

−

(

0.04

×

6 )

)^

7

/

10=

t

−

4.7619

or

t

=

0.832

h

r

s

6 0

3 years ago