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3 years ago

What force impacts all machines and reduces its energy?

Mathematics

2 answers:

omeli [17]3 years ago

5 0

I think it’s frictional force sorry if it’s wron g

Contact [7]3 years ago

3 0

Answer:

Step-by-step explanation:

FRICTION impacts all machines and reduces energy.

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Estimates show that there are 1.4x10^8 pet fish and 9.4x10^6 pet reptiles in the United States. How many pet fish and reptile pe

azamat

Answer:

There are total number of pet fish and reptile pets are $1.49\times 10^8$ .

Step-by-step explanation:

Number of pet fish are $1.4\times 10^8$ and the number of pet reptiles are $9.4\times 10^6$ .

It is required to find the total number of pet fish and reptile pets in the United States. It can be done by simply adding the above numbers such that total numbers are given by :

$T=1.4\times10^{8}+9.4\times10^{6}\\\\T=149400000\\\\T=1.49\times 10^8$

So, there are total number of pet fish and reptile pets are $1.49\times 10^8$ .

8 0

3 years ago

Triangle ABC is an isosceles triangle. The length of the base is 20 meters and the corresponding height is 24 meters. Find the p

olga nikolaevna [1]

Answer:

72, I believe.

8 0

4 years ago

A company that makes wildlife videos purchases camera equipment for $32,400. The equipment depreciates in value at a constant ra

alexdok [17]

Answer:

Step-by-step explanation:

yearly depreciation

32,400÷12

=2,700

Accumulated depreciation for 4 years

2,700×4

=10,800

Book value at year 4

32,400−10,800

=21,600

3 0

3 years ago

Read 2 more answers

A 10-foot ladder leans against a wall so that it is 6 feet high at the top. The ladder is moved so that the base of the ladder t

konstantin123 [22]

Answer:

The top of the ladder is now at 10 ft.

Step-by-step explanation:

At the start, we have a height H=6, a length L=10 and a base B, that has to be calculated by the Pythagorean theorem:

$B^2=L^2-H^2=10^2-6^2=100-36=64\\\\B=\sqrt{64}=8$

The base is moved twice the distance the height moves up.

We called this distance x, so we have:

$L^2=(H+x)^2+(B-2x)^2=H^2+2Hx+x^2+B^2-4Bx+4x^2\\\\L^2=(H^2+B^2)+5x^2+(2H-4B)x\\\\L^2=L^2+5x^2+(2H-4B)x\\\\0=5x^2+(2H-4B)x\\\\5x+(2H-4B)=0\\\\x=\dfrac{4B-2H}{5}=\dfrac{4*8-2*6}{5}=\dfrac{32-12}{5}=\dfrac{20}{5}=4$