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Sever21 [200]
3 years ago
10

Please help

Physics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

I'm pretty sure that the answer is A

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the royal Gorge Bridge in Colorado rises 321 m above the Arkansas river. suppose you kick a rock horizontally off the bridge. Th
KengaRu [80]

Answer:

2.48 m/s

Explanation:

We can use the kinematic equation,

s = ut +½at²

Where

s = displacement

u = initial velocity

t = time taken

a = acceleration

Using the equation in vertical direction,

321 = 0×t +½×g×t², u = 0 because initial vertical velocity is 0

We get t = 8.01 s

Using the equation in the horizontal direction,

52 = u×8.01 +½×0×(8.01)²,. a = 0 because no unbalanced force act on object in that direction

So u = 2.48 m/s

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3 years ago
If the period of a wave is 0.05 s, then what is its frequency?
yKpoI14uk [10]
Frequency = 1/time period = 1/0.05 = 20s^-1. 
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Who wrote the universal gravitation
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I think it was Isaac Newton
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allochka39001 [22]
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8 0
3 years ago
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A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista
slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

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

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

C' = \frac{1}{2}C

The potential difference across the capacitor is given by

V= \frac{Q}{C}

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

V' = 2 V

Now we can analyze the electric field between the plates of the capacitor, which is given by

E=\frac{V}{d}

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

E'=\frac{2V}{2d}=\frac{V}{d}=E

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0
3 years ago
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