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2 years ago

PLEASE HELP!!!! What are 3 ways that machines can make doing work easier? (thats the whole question)

Physics

2 answers:

otez555 [7]2 years ago

7 0

Answer:

they can do work easier for an example a car a car make things easier by going some were far it can also be used to help people travel instead of a plane it can also be used for carriying things like a boat. (you can put it in a diffrent way)

Explanation:

hope it helps! :)

Rudiy272 years ago

4 0

Answer:

Machines reduce the time of work hence reducing the rate of doing work ( power ).

Machines e.g pulleys carry heavy loads with a less and reasonable effort.

Machines e.g generators induce current in a limited amount of time

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Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

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