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Physics

2 answers:

mel-nik [20]3 years ago

7 0

The rule for the sequence is: You add 4/6 each time
5/6 +4/6 = 1 1/2
1 1/2 + 4/6 = 2 1/6
2 1/6 +4/6 = 2 5/6

muminat3 years ago

3 0

The rule is that 4/6 is added each time, and when adding or subtracting a fraction the denominator stays the same

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An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

3 years ago

Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of

Andre45 [30]

Well first of all, I think the students may have been correct.
If they didn't use distilled water, and if it wasn't exactly at
standard temperature, then the mass of 25.0 mL could
very well be 25.4 grams. We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

+0.4 / 25.0 = +0.016 .

To change a decimal to a percent, move the
decimal point two places that way ===> .

+ 0.016 = +1.6 % .

Their measurement was 1.6% too high.

Let's not call it an 'error'. Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value. OK ?

4 0

3 years ago

Read 2 more answers

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SVETLANKA909090 [29]

Answer:

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3 years ago

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