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3 years ago

How does science help society?

Physics

2 answers:

Arada [10]3 years ago

8 0

Learn and create new things. It also builds our economy

Ber [7]3 years ago

3 0

It helps us learn more about the world around us

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What is difference between heat engine and carnot engine

labwork [276]

Answer: A heat engine uses temperature differences which cause pressure changes to exert force on a moving part. A Carnot Process is a theoretical explanation of a process involving pressure and temperature changes during ,amongst other things, phase changes.

Explanation:

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1 year ago

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a body starts from rest and accelerates uniformly at 5ms‐2. Calculate the time taken by the body to cover a distance of 1km

mixas84 [53]

The body will take 20 seconds to cover a distance of 1000 m i.e. 1 km

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3 years ago

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7. A golfer hits a golf ball a through a horizontal displacement of 80m. The time of flight for the ball was 1.8

Darya [45]

Answer:

44.4 m/s

Explanation:

d = 80 m, t = 1.8 s, find v

d = v*t

v = d/t = 80/1.8 = 44.4 m/s

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2 years ago

Any one please give me the correct answer of this question???<br> I hope you can help me...

Phantasy [73]

Answer:

no one ever amswers my questions

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago