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Radda [10]
3 years ago
13

What is an equilibrant

Physics
1 answer:
professor190 [17]3 years ago
5 0
The question seems to be what is an equilibrant force.

The answer is "an added force that produces equilibrium.

Here you have more insight:

<span>an object that has no net force acting on it? This object indeed is in equilibrium but the object is not the equilibran force.

the reaction force in an action-reaction pair of forces?

the reaction force is not an equilibrant force. The reaction force exists always but equilibrium is only possible if the net force is cero.

an added force that produces equilibrium? this is the right  answer.</span>
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A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m
saw5 [17]

Answer:

Ep = 3924 [J]

Explanation:

To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

E_{p}=m*g*h\\

where:

Ep = potential energy [J] (units of Joules)

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 10 [m]

E_{p} =40*9.81*10\\E_{p} = 3924 [J]

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photoshop1234 [79]

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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
A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?
Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

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a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0
3 years ago
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