Can anyone help me with this problem? It is -2.1??

In a pulley system, a 5-newton weight is to be lifted 2 meters. The rope is pulled 10 meters. The input force is two newtons.

loris [4]

The answers to the problem are as follows:

MA= 5

IMA= input distance/ output distance, 5

<span>AMA= output force/ input force, 5/2
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

2 years ago

Read 2 more answers

A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected ac

Sonbull [250]

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

$\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V$

The rms voltage (in V) measured across the secondary coil is calculated as;

$V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V$

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

7 0

2 years ago

Match each word to it's correct meaning.

Ivan

Answer:

fluid transfer of heat by the motion of the particles of gas or fluid

8 0

1 year ago

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What could you do to change the volume of a gas?

Lorico [155]

The ONLY way to change the volume of a sample of gas is to transfer it to a container with different volume.
Simply changing its temperature or pressure in the same jar won't do it. Any amount of gas always fills whatever container you keep it in.

8 0

2 years ago

a mass of 1.00 kg of water at temperature T is poured from a height of 0.100 km into a vessel containing water of the same tempe

Mariana [72]

Answer:

1.34352 kg

Explanation:

$m_w$ = Mass of water falling = 1 kg

h = Height of fall = 0.1 km

$\Delta T$ = Change in temperature = 0.1

c = Specific heat of water = 4186 J/kg K

g = Acceleration due to gravity = 9.81 m/s²

$m_v$ = Mass of water in the vessel

Here the potential energy will balance the internal energy

$m_wgh=m_wc\Delta T+m_vc\Delta T\\\Rightarrow m_v=\dfrac{m_wgh-m_wc\Delta T}{c\Delta T}\\\Rightarrow m_v=\dfrac{m_wgh}{c\Delta T}-m_w\\\Rightarrow m_v=\dfrac{1\times 9.81\times 100}{4186\times 0.1}-1\\\Rightarrow m_v=1.34352\ kg$

Mass of the water in the vessel is 1.34352 kg

6 0

2 years ago