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Dennis_Churaev [7]
3 years ago
13

Which of the following can be reduced to a single number in standard form?

Physics
1 answer:
raketka [301]3 years ago
6 0

Complete question is;

Which of the following can be reduced to a single number in standard form?

A) 3√3 + 5√8

B) 2√5 + 5√45

C) √7 + √9

D) 4√2 + 3√6

Answer:

Only option B) 2√5 + 5√45 can be reduced to a single number

Explanation:

A) For 3√3 + 5√8;

Let's simplify it to get;

3√3 + 5√(4 × 2)

From this, we get;

3√3 + (5 × 2)√2 = 3√3 + 10√2

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

B) 2√5 + 5√45

Simplifying to get;

2√5 + 5√(9 × 5)

This gives;

2√5 + (5 × 3)√5 = 2√5 + 15√5

Adding the surds gives;

17√5.

This is a single number and thus can be reduced to a single number

C) For √7 + √9

Simplifying, to get;

√7 + 3.

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

D) 4√2 + 3√6

Thus can't be simplified further because both numbers inside the square root don't have factors that are perfect squares.

Thus, it remains 2 numbers and not a single number and can't be reduced to a single number in standard form.

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A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th
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Answer:

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Explanation:

The equation that relates heat Q with the temperature change T-T_0 of a substance of mass <em>m </em>and specific heat <em>c </em>is Q=mc(T-T_0).

We want to calculate the final temperature <em>T, </em>so we have:

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Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 273^{\circ} to our temperature in ^{\circ}C to have it in ^{\circ}K as it must be):

T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K

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what particle moves from one object to another so that the object is either positively charged or negatively charged
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A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during
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Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

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Using the II equations of motion,

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Substituting the given values,

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