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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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Its A i did it and i got a
n200080 [17]

Answer:

THATS COOL

Explanation:

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3 years ago
Fill in the appropriate values for each blank as it refers to ATOM 1. The number of protons present in ATOM 1 is _________.​
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3, protons are positive and there are 3 positive atoms visible
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3 years ago
With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

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7 0
1 year ago
calculate the amount of work done by a person while taking a bag of mass 100kg to the top of the building hight 10m. The mass of
vredina [299]

Explanation:

Total mass=100+10=110

Total weight=mass×gravitational field strength

=110×10

=1100N

Work done=force×distance

=1100×10

=11000J

<em>Please mark me as brainliest if this helped you!</em>

6 0
2 years ago
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b
Radda [10]

Answer:

\theta=34 \textdegree

Explanation:

From the question we are told that:

Mass m=55kg

Angle \theta =28.0

Coefficient of static friction \alpha =0.680

Generally, the equation for Newtons second Law is mathematically given by

For

\sum_y=0

N=mgcos \theta

for

\sum_x=0

F_{s}=mgsin\theta

Where

F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta

F_{s}=0.68*55*9.8*cos 28

F_{s}=323.62N

Therefore

\alpha mgcos \theta=mg sin \theta

\theta=tan^{-1}(0.68)

\theta=34 \textdegree

6 0
2 years ago
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