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seraphim [82]
3 years ago
7

How will you express the following in m: a. 10 nm b. 110 pm c. 120 mm d. 200 km Please help me

Physics
1 answer:
7nadin3 [17]3 years ago
4 0

Answer:

1 km = 1000 m

1 m = 100 cm

1 m = 1000 mm

1 cm = 10 mm

hope this helps u figure it out :)

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A jewellery melts 500g of Silver to pour into a mould. Calculate how much energy was released as the silver solidified.
irga5000 [103]

When silver is poured into the mould the it will solidify

In this process the phase of the Silver block will change from liquid to solid.

This phase change will lead to release in heat and this heat is known as latent heat of fusion.

The formula to find the latent heat of fusion is given as

Q = mL

here given that

m = mass = 500 g

L = 111 kJ/kg

now we can find the heat released

Q = 0.5 * 111 kJ

Q = 55.5 kJ

So it will release total heat of 55.5 kJ when it will solidify

8 0
3 years ago
What is the smallest accurate measurement that can be made with an instrument?
vovangra [49]
Dependent on what you are measuring and what took you are using. Please be more specific.
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Maurice pulls on the end of a spring scale. He lets go of the end and observes the spring snap back into place. What force resto
andreev551 [17]
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3 0
3 years ago
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A two-slit Fraunhofer interference-diffraction pattern is observed with light of wavelength 672 nm. The slits have widths of 0.0
ololo11 [35]

Answer:

Explanation:

In case of diffraction , angular width of central maxima =2 λ/d

λ is wave length of light and d is slit width

In case of interference , angular width of each fringe

= λ /D

D is distance between two slits

No of interference fringe in central diffraction fringe

=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.

6 0
3 years ago
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
3 years ago
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