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Andre45 [30]
3 years ago
6

A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in met

ric tonnes per cubic metre? Give your answer to 1 significant figure.​
Physics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

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The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the
natta225 [31]

Answer:

50 N

Explanation:

Efficiency of a machine can't be more than 1, so I assume you mean 40%.  (Remember, efficiency and mechanical advantage are not the same).

Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

We know that e = 0.40, and Fout = 120 N.  Since there are 6 pulleys, we also know that Din/Dout = 6.

F = (120 N / 0.4) / 6

F = 50 N

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2 years ago
Which activity demonstrates the lowest level of intensity? sprinting to catch a bus lying on the couch playing video games walki
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<span>lying on the couch


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8 0
3 years ago
Read 2 more answers
How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 4.00
klio [65]

Explanation:

Let us assume that the separation of plate be equal to d and the area of plates is 9 \times 10^{-4} m^{2}. As the capacitance of capacitor is given as follows.

            C = \frac{\epsilon_{o}A}{d}

It is known that the dielectric strength of air is as follows.

               E = 3 \times 10^{6} V/m

Expression for maximum potential difference is that the capacitor can with stand is as follows.

                       dV = E × d

And, maximum charge that can be placed on the capacitor is as follows.

               Q = CV

                   = \frac{\epsilon_{o} A}{d} \times E \times d

                   = \epsilon_{o}AE

                   = 8.85 \times 10^{-12} \times 3 \times 10^{6} \times 4 \times 10^{-4}

                   = 1.062 \times 10^{-8} C

or,                = 10.62 nC

Thus, we can conclude that charge on capacitor is 10.62 nC.

5 0
3 years ago
Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 10^8/ (this is the speed of light in vacuu
zlopas [31]

Answer:

Explanation:

we know that

s=vt here v is the speed and s is distance covered by the signals

given data

v=3*10^8

t=10 min we have to convert it into seconds

1 minute=60 seconds

so

10 minutes =10*60/1 =600 seconds

now putting the value of v and t we can find the value of s

s=vt

s=3*10^8*600

s=1.8*10^11m

i hope this will help you

8 0
3 years ago
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