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jek_recluse [69]
3 years ago
14

The Robinson projection map is considered very useful because

Physics
1 answer:
Vladimir79 [104]3 years ago
7 0
Hello,
   The question states: <span>The Robinson projection map is considered very useful because...
The answer is \boxed{because \ most \ distances, \ sizes, \ and \ shapes \ are \ accurate.}

Hope this helped!

~FoodJunky
</span>
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HELP!!!!! Please hurry up
Degger [83]

it's def. TRUE. i got the same question and i got it right

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3 years ago
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Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and
sleet_krkn [62]

Answer:

\alpha=78.4\ rad.s^{-2}

Explanation:

Given:

  • mass of solid cylinder, m=50\ kg
  • diameter of cylinder, d=0.25\ m
  • mass of bucket of water, m_w=20\ kg

<em>When the bucket is released to fall in the well, it fall under the acceleration due to gravity.</em>

We have formula for angular acceleration as:

\alpha=\frac{g}{r}

where:

g = acceleration due to gravity

r = radius of the cylinder

\aplha=\frac{9.8}{0.125}

\alpha=78.4\ rad.s^{-2}

4 0
3 years ago
When talking about variables in a scientific experiment, describe how you know what the independent variable, dependent variable
Mnenie [13.5K]

An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.

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3 years ago
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A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a
yawa3891 [41]
Write out what you have which is:
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6 0
3 years ago
A charge of 6.7 × 10^-15 coulombs is located at a point where its potential energy is 5.6 × 10^-12 joules. What is the electric
natita [175]

Answer:

C. 8.4 × 10^2 volts

Explanation:

The potential energy of a charge is given by:

U=qV

where

q is the magnitude of the charge

V is the electric potential

In this problem, we have

q=6.7\cdot 10^{-15} C is the charge

U=5.6\cdot 10^{-12} J is the potential energy

Re-arranging the formula and using these numbers, we can find the electric potential:

V=\frac{U}{q}=\frac{5.6\cdot 10^{-12} J}{6.7\cdot 10^{-15} C}=835.8 V = 8.4\cdot 10^2 V

4 0
3 years ago
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