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4 years ago

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An object falls 210 m in 6.3 s. What was it's initial velocity? Express your answer with the correct number of significant digit

s.

1 answer:

KIM [24]4 years ago

8 0

Answer:

$u = 2.5 \ m/s$

Explanation:

Given data:

Height, $h = 210 \ m$

Time, $t = 6.3 \ s$

Let the initial velocity of the object be <em>u</em>.

From the kinematic equation,

$h = ut + \frac{1}{2}gt^{2}$

$210 = u \times 6.3 + 0.5 \times 9.80 \times (6.3)^{2}$

$\Rightarrow \ u = 2.46 \ m/s$

$u = 2.5 \ m/s$ (rounding to tenth place)

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Read 2 more answers

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

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Answer:

The change in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

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From the question we are told that

The magnitude of the uniform electric field is $E = 950 \ N/C$

The distance traveled by the electron is $x = 2.50 \ m$

Generally the force on this electron is mathematically represented as

$F = qE$

Where F is the force and q is the charge on the electron which is a constant value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally the work energy theorem can be mathematically represented as

$W = \Delta KE$

Where W is the workdone on the electron by the Electric field and $\Delta KE$ is the change in kinetic energy

Also workdone on the electron can also be represented as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ considering that the movement of the electron is along the x-axis

So

$\Delta KE = F * x cos (0)$

substituting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

Now From the law of energy conservation

$\Delta PE = - \Delta KE$

Where $\Delta PE$ is the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

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3 years ago