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3 years ago

A sensational 130 kg midfielder has 695 J of

Physics

1 answer:

antoniya [11.8K]3 years ago

4 0

Answer:

v = 3.27 m/s

Explanation:

KE = 1/2 mv^2

695 J = 1/2 (130kg)(v^2)

695 J / (1/2 x 130kg) = v^2

v^2 = square root of 10.69

v = 3.27 m/s

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For the scenarios in which total internal reflection is possible, rank the scenarios on the basis of the critical angle, the ang

likoan [24]

Answer:

Explanation:

For critical angle C the relation is

SinC = 1 / μ where μ is refractive index of denser medium with respect to rarer medium .

In case of A , refractive index of denser medium with respect to rarer medium .is equal to 1.33 / 1 = 1.33

In case of B , refractive index of denser medium with respect to rarer medium .is equal to 2.42 / 1 = 2.42

In case of C , refractive index of denser medium with respect to rarer medium .is equal to 2.42 / 1.33 = 1.82

In case of D , refractive index of denser medium with respect to rarer medium .is equal to 1.50 / 1.33 = 1.127

In case of D , refractive index is lowest , critical angle will be highest .

In case of B , refractive index of denser medium with respect to rarer medium .is highest so critical angle will be lowest.

Ranking from largest to smallest

D > A >C >B .

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3 years ago

If a Man suit a spare at a place where

Mazyrski [523]

Answer:

Yes

Explanation:

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3 years ago

Which of the following items has the most inertia while at rest?

Vika [28.1K]

The answer should be (c) because the dog is the one which make a sound of ooooooo

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3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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3 years ago

TIMED! URGENT! REALLY APPRECIATE HELP!! TYSM!!!!!!

Diano4ka-milaya [45]

The answer is 27.03 I just multiplied the two numbers

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3 years ago

Read 2 more answers