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Nastasia [14]
3 years ago
9

a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel d

uring 30 seconds of rotation?​
Physics
1 answer:
vlabodo [156]3 years ago
8 0

Answer: 50π m ≈ 157 m

Explanation:

100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s

d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m

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If Sarah melts 10g of tin. What mass of melted tin would she have at the end of the experiment?​
Nastasia [14]

Answer:

10g

Explanation:

As the Law of Conservation of Mass states that " Mass can neither be created nor be destroyed in a chemical reaction".

Though melting of tin isn't a chemical change, the same logic is applied here...

Hence,

The mass of tin will be 10 g itself...

7 0
3 years ago
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Match the measurements to the correct number of significant figures.
Shalnov [3]
To calculate the number of sig figs
1.  Count ALL nonzero numbers
2.  If the zero is between 2 numbers, count it
3.  If in a decimal the zero is at the end, count it
4.  In a decimal all the zeros before the first nonzero number are placeholders and don't count them
5.  In a number greater than zero all zeros AFTER the last nonzero number are placeholders and don't count them.

A - 5
B - 2
C - 3
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8 0
2 years ago
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Which of the following is an electromagnetic wave?
morpeh [17]
Your answer will be Radio Waves . 

That seems to be the only to make sense. Hope that helps u 
5 0
3 years ago
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In general, which statement does your data support ?
hjlf

Answer:

A

Explanation:

As distance increases, velocity increases.

5 0
2 years ago
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition
Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So mg=Kx

K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

Now we know that \omega =\sqrt{\frac{K}{m}}

So \frac{2\pi }{T} =\sqrt{\frac{K}{m}}

\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

T=2.1371sec

8 0
3 years ago
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