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2 years ago

if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

1 answer:

6 0

Since the stone was dropped from height, its initial velocity = 0 m/s

Using v² = u² + 2gs.

Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s

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2 years ago

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grandymaker [24]

Answer:

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3 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

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3 years ago

A biologist looking through a microscope sees a bacterium at r⃗ 1=2.2i^+3.7j^−1.2k^μm(1μm=10−6m). After 6.2 s , it's at r⃗ 2=4.6

stiv31 [10]

The Average velocity for the bacterium is 0.75 unit/sec.

<u>Explanation:</u>

The given values are in the vector form

Where,

dS = distance covered

dT = time interval

Now, to calculate distance covered, we have

$|d S|=\sqrt{d S^{2}}$

$d S=r_{2}-r_{1}$

d S=(4.6 i+1.9 k)-(2.2 i+3.7 j - 1.2 k)

d S=(4.6-2.2) i+(0-3.7) j+(1.9+1.2) k

d S=2.4 i-3.7 j+3.1 k

Now, putting these values in the standard formula to evaluate the average velocity, we get;

$v_{a v g}=\frac{|\mathrm{d} S|}{d T}$

$v(a v g)=\frac{|\sqrt{\left\{\left(2.4^{2}\right)+\left(3.7^{2}\right)+\left(3.1^{2}\right)\right\}}|}{7.2}$

As dT=7.2 sec

Now,

Solving the equation, we get;

$v(a v g)=\frac{5.390732789}{7.2}$

$\begin{aligned}&v(a v g)=\frac{5.39}{7.2}\\&v(a v g)=0.748611111\\&v(a v g)=0.75 \text { units / sec }\end{aligned}$

Hence, the average velocity for the bacterium is 0.75 unit/sec.

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3 years ago

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Answer:

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3 years ago