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nikitadnepr [17]
3 years ago
9

A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t

he rope to a height of 33 meters
Physics
1 answer:
Kazeer [188]3 years ago
6 0

Answer:

Explanation:

Given

Length of rope L=55\ m

Weight of rope W=4040\ N

weight density \lambda =\frac{4040}{55}=73.45\ N/m

Work done to lift rope 33 m

W=\int_{0}^{33}\lambda hdh

W=\int_{0}^{33}73.45hdh

W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0

W=39.993\ kJ      

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What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k
Lyrx [107]

We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

\sf \: F=ma

Where,

  • F is force
  • m is mass
  • a is acceleration

In our case,

  • F = ?
  • m = 2500 kg
  • a = 20m/s

\tt \: F_{net}  = 2500 \times 20 \\   \tt= 50000

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>

3 0
3 years ago
Read 2 more answers
A rectangular piece of​ cardboard, whose area is 352 square​ centimeters, is made into an open box by cutting a 2​-centimeter sq
Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

xy=352

y=\dfrac{352}{x} -------- (1)

The volume of the rectangular piece

2(x-4)(\dfrac{352}{x}-4)=432

x^2-38x+352=0

(x-16)(x-22)=0

x=16,22

Put the value of x in the equation (I)

For x = 16

y=\dfrac{352}{16}=22

For x = 22

y=\dfrac{352}{22}=16

Dimension of cardboard is 22 m by 16 m

8 0
3 years ago
How do theories differ from Laws and Principles?
g100num [7]
A scientific law is the simple mathematical expression of the relationship involved. A principle is the same relationship expressed in words. A theory is the explanation of the facts that make up the relationship.
8 0
3 years ago
A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
krek1111 [17]

Answer:

a) Pb= 200 PA

b).work done= -3600 joules

c).3600joules

D).the system works under isothermal condition so no heat was transferred

Explanation:

2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.

a). PbVb= PaVa

Pb= (PaVa)/VB

Pb= (600*3)/9

Pb= 1800/9

Pb= 200 PA

b). work done= n(Pb-Pa)(Vb-Va)

Work done= 2*(200-600)(9-3)

Work done= -600(6)

Work done=- 3600 Pam³

work done= -3600 joules

C). Change in internal energy I the work done on the system

= 3600joules

D).the system works under isothermal condition so no heat was transferred

4 0
3 years ago
Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
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