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3 years ago

10

Which element is most likely to be shiny?

2 answers:

mestny [16]3 years ago

5 0

Answer:

Calcium, or as we intellectuals call it, Bone Juice

Explanation:

stellarik [79]3 years ago

4 0

calcium (ca) is most likely to be shiny

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An object is placed 10 cm in front of a diverging mirror. What is the focal length of the mirror if the image appears 2 cm behin

Dafna1 [17]

Answer:

the focal length of the mirror is : $f=-2.5\,\,cm$

Explanation:

Use the formula for the formation of image using a divergent mirror and recalling that the image (s') that this mirror formed is virtual, so it is entered as a negative number in the formula. Use the object position (s) as 10, the image position (s') as -2, and derive the value of the focal length:

$\frac{1}{s} +\frac{1}{s'}=\frac{1}{f}\\\frac{1}{10} +\frac{1}{-2}=\frac{1}{f}\\\frac{1}{10} -\frac{1}{2}=\frac{1}{f}\\\frac{10\,f}{10} -\frac{10\,f}{2}=\frac{10\,f}{f}\\f-5\,f=10\\-4\,f=10\\f=-2.5\,\,cm$

6 0

3 years ago

What’s is the movement of one object around another

allsm [11]

Answer:

revolution

Explanation:

8 0

3 years ago

Describe the role of minerals in the formation of rocks

klio [65]

<span>Minerals make up rocks. The role of minerals in rock formation is largely dependent on how the rock is formed.</span>

3 0

3 years ago

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

a load of 400 Newton is lifted by a first class lever in which the load is at the distance of 20 cm and the effort is at the dis

Len [333]

Answer:

solution,
Given
load =400N
ld=0.2m
ed=0.6m
effort =150N

Explanation:

efficiency =output work/input work ×100%

l×ld/e×ed×100%

400×0.2/150×0.6×100%

80/90×100%

88.89%ans

7 0

2 years ago