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3 years ago

If the sum of the external forces on an object is zero, then the sum of the external torques on it

Physics

1 answer:

Zarrin [17]3 years ago

5 0

Answer:

True.

Explanation:

If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.

The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.

Hence, the given statement is true.

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Read 2 more answers

There are four springs stretched by the same mass.

brilliants [131]

Spring C stretches 100 cm.

Explanation:

The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.

Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.

According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"

F = ke

k is the spring constant

e is the extension

We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.

Learn more;

Force brainly.com/question/8882476

#learnwithBrainly

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2 years ago

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

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

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

$Q = 1.245 \times 10^{-12}$

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3 years ago