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MariettaO [177]
4 years ago
11

Using a simple machine, the machine cannot increase the amount of _______ , but can increase the amount of ________ .

Physics
2 answers:
konstantin123 [22]4 years ago
5 0

Answer:

work, force

Explanation:

givi [52]4 years ago
4 0

work but not energy

Explanation:

hdhhsnshshshgsnajsysvbskdk

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What is the wavelength of a wave?
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A 1200N load is to be lifted with 200N effort using a first class lever. At what distance
kipiarov [429]

Explanation:

Hey there!!

Let's simply work with it.

Here,

load = 1200N

Effort = 200N

Load distance = 15cm

We have,

According to the principle of lever.

L×LD = E×ED.

1200×15 = 200× ED.

18000 = 200ED.

ed =  \frac{18000}{200}

Therefore, Effort Distance = 90cm.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

7 0
3 years ago
At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of banan
Elina [12.6K]

Answer:

2.67kg

Explanation:

The maximum velocity, v _ {max} of a body experiencing simple harmonic motion is given by equation (1);

v_{max}=\omega A............(1)

where \omega is the angular velocity and A is the amplitude.

The problem describes the oscillation of a loaded spring, and for a loaded spring the angular velocity is given by equation (2);

\omega=\sqrt{\frac{k}{m}}.................(2)

where k is the force constant of the spring and m is the loaded mass.

We can make \omega the subject of formula in equation (1) as follows;

\omega=\frac{v_{max}}{A}.................(3)

We then combine equations (2) and (3) as follows;

\frac{v_{max}}{A}=\sqrt{\frac{k}{m}}.................(4)

According to the problem, the following  are given;

v_ {max }=1.92m/s\\A=0.21m\\k=223N/m

We then substitute these values into equation (4) and solve for the unknown mass m as follows;

\frac{1.92}{0.21}=\sqrt{\frac{223}{m}}

9.143=\sqrt{\frac{223}{m}}

Squaring both sides, we obtain the following;

9.143^2=\frac{223}{m}\\9.143^2*m=223\\83.592m=223\\therefore\\m=\frac{223}{83.592}\\m=2.67kg

8 0
3 years ago
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