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Andrew [12]
3 years ago
8

Physics 1 Course. please help. ​

Physics
1 answer:
Aleks [24]3 years ago
6 0

Answer:

a. 1.64 m/s²

Explanation:

Centripetal acceleration is the square of tangential velocity divided by the radius.

a = v²/r

First, convert km/h to m/s.

30.0 km/h (1000 m/km) (1 h / 3600 s) = 8.33 m/s

Find the acceleration.

a = (8.33 m/s)² / (42.4 m)

a = 1.64 m/s²

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Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astr
Aloiza [94]

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

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7 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

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VashaNatasha [74]
YES I TOOK THIS ! The answer is 25.79.
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