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likoan [24]
3 years ago
10

A body of mass 80kg was lifted vertically through a distance of 5.0 metres. Calculate the work done on the body. ( Acceleration

due to gravity g=10ms²)​

Mathematics
2 answers:
tamaranim1 [39]3 years ago
4 0

Answer:

80×5×10=4000J

so therefore, the work done on the body was 4000J

Murljashka [212]3 years ago
3 0

Answer:

4000N

Step-by-step explanation:

Check the attachment

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Brad has eight more nickels than quarters. Altogether he has $6.10. how many of each does he have
wlad13 [49]

Brad have 19 quarters and 27 nickels.

Step-by-step explanation:

Given,

Total worth of nickels and quarters = $6.10 = 6.10*100 = 610 cents

One quarter = 25 cents

One nickel = 5 cents

Let,

x represent the number of quarters.

y represent the number of nickels.

According to given statement;

25x+5y=610     Eqn 1

y = x+8              Eqn 2

Putting value of y from Eqn 1 in Eqn 2

25x+5(x+8)=610\\25x+5x+40=610\\30x=610-40\\30x=570

Dividing both sides by 30

\frac{30x}{30}=\frac{570}{30}\\x=19

Putting x=19 in Eqn 2

y=19+8\\y=27

Brad have 19 quarters and 27 nickels.

Keywords: linear equation, subtraction

Learn more about subtraction at:

  • brainly.com/question/9369548
  • brainly.com/question/9328925

#LearnwithBrainly

7 0
3 years ago
Over 5 days 5,385 people went to Chick-Fil-A. If the same amount of people went each day, how many people went on one day?
Naily [24]

Answer:

1077

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Jensen spends $18.00 for every 72 text messages she sends. How many messages can Jensen send for $1.00
Rasek [7]
4 texts because 18 dollars for 72 texts
can reduce to 1 dollar for 4 texts. i can try to explain further if needed
8 0
3 years ago
A circle is centered at $O$ and has an area of $48 \pi.$ Let $Q$ and $R$ be points on the circle, and let $P$ be the circumcente
nikklg [1K]

Answer:

Area of Triangle QRP = 3\sqrt{3}

Step-by-step explanation:

According to Question , We have a circle With Centre 'O' & Area 48\pi .

Area Of Circle = 48\pi

\pir^{2} = 48\pi

r = \sqrt{48}

Now We Have Two Points Given On Circle Q & R , P Is Circumcentre Of Triangle QRO .

Thus A Circle Can Also Be Formed with Centre P . ( See attachment For Diagram )

Now The Diameter of Circle With Centre P = Radius Of Circle with Centre O

so Radius Of Circle With Centre P(r_{2}) = \frac{\sqrt{48}}{2}

Now We Have To Find Area Of Equilateral Triangle .

A = \frac{\sqrt{3}}{4} r_{2} ^{2}

A= \frac{\sqrt{3} }{4} * \frac{\sqrt{48} }{2}*\frac{\sqrt{48} }{2}

The Area Of PQR is = 3\sqrt{3}

For Diagram , Please Find In Attachment

7 0
3 years ago
Which of the following is NOT true about a rhombus?
Naily [24]

Answer:

Which of the following is NOT true about a rhombus?

options:

lines of symmetry are the perpendicular bisectors

lines of symmetry are the diagonals

a rhombus has four congruent sides

a rhombus is a quadrilateral

5 0
3 years ago
Read 2 more answers
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