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sdas [7]
3 years ago
6

The sum of all forces and components in both x and y are equal to zero, as well as in the sum of all moments of force or torsion

generated by external external forces from a point P in the plane or outside This is the sound of this body in:
Select one:
to. Movement
second. Balance
do. Reacting
re. Dynamic balance
Physics
1 answer:
12345 [234]3 years ago
7 0
1,2,3,4 is the answer, your welcomethe answer is movement !
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The initial volume reading in a graduated cylinder is 30 mL. The mass of an irregular shape of an unknown metal piece is 55.3 g.
Vadim26 [7]

Answer:

7.9\frac{gr}{cm^3}

Explanation:

When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.

So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7cm^3).

Density is \frac{mass}{volume}.

So the density of that piece of metal is \frac{55.3g}{7cm^3}

Which leaves us with a final density of 7.9\frac{gr}{cm^3}

6 0
3 years ago
Do you divide mass by volume to get density
bogdanovich [222]
Density<span> is the </span>mass<span> of an object </span>divided<span> by its </span>volume<span>. So the answer would be Yes. Hope it helps! (:</span>
7 0
3 years ago
Why does an object above the ground have P.E.?
Advocard [28]
This doesn't make sense
6 0
3 years ago
Read 2 more answers
Which uses motion of water in nature to produce electricity?
Zinaida [17]

Answer: D. Hydroelectric power

Explanation: Hydroelectric means water power (not exactly but "hydro" is water and electric is well self explanatory)

6 0
3 years ago
A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S=\frac{360}{10000}=0.036 m^2

\Delta d=0.8 cm=0.008 m

\Delta V=100 V

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

C=\frac{\epsilon_0 S}{d}

Capacitance of capacitor after moving plates

C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

Potential difference between plates after moving

V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}

\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100

\frac{Q\Delta d}{\epsilon_0 S}=100

\epsilon_0=8.85\times 10^{-12}

Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}

Q=3.9825\times 10^{-9} C

Hence, the charge on positive plate of capacitor=Q=3.9825\times 10^{-9} C

6 0
3 years ago
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