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lozanna [386]
3 years ago
12

A ball is dropped from a height of 20m. If its velocity increases uniformly at the rate 10m/s2 with what velocity and after what

time will it strike the ground?
Physics
2 answers:
belka [17]3 years ago
6 0

Answer: 2 seconds

Explanation:

adelina 88 [10]3 years ago
3 0

Since I've only memorized a couple of formulas for these things, I need to do the time first:

I'll use the formula  Distance = (1/2) (acceleration) (time)²

20 m = (1/2) (10 m/s²) (time)²

Time² = (20 m) / (5 m/s²)

Time² = 4 sec²

<em>Time = 2 seconds</em>

Now ...

-- When the ball hits the ground, it has been falling for 2 seconds.

-- Its acceleration has caused its speed to increase by 10 m/s every second.

-- So after 2 seconds, its speed has grown to (2 s) (10 m/s²) = <em>20 m/s</em> .

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A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential
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Explanation:

Given that,

Electric field = 5750 N/C

Charge q=+10.5\times10^{-6}\ C

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(a). When the charge is moved in the positive x- direction

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Using formula of electric potential energy

\Delta U=-W

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\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}

\Delta U=-3.32\times10^{-3}\ J

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Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot (-d)

\Delta U=-q(E\cdot (-d))

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})

\Delta U=3.32\times10^{-3}\ J

The change in electric potential energy  is 3.32\times10^{-3}\ J

Hence, This is the required solution.

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