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2 years ago

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Is v2 = v1t+a dimensionally correct? Explain please!

1 answer:

Lady bird [3.3K]2 years ago

4 0

You want v2 = v1 + at
v is measured in m/s, a in m/s2, and t in s.
the dimensions multiply like algebraic quantities.
so because v2 is measured in m/s, then (v1 + at) has to come out in m/s
the units for (v1 + at) are (m/s) + (m/s2)(s)
time "s" cancels out one acceleration "s", so it comes ut to (m/s) + (m/s), which = (m/s).
if you had (v1t + a), then you would have (m/s)(s) + (m/s2) which = (m) + (m/s2), which doesn't work.

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valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

$K=\frac{1}{2}I \omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

$K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2$

For disk 2:

$K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2$

so the total energy is

$K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2$

So, none of the options is correct.

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2 years ago

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

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$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

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2 years ago

An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v

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We are given: Final velocity ( $v_f$ )=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

$v_f = v_i + at.$

Let us plug values of final velocity, and time in above equation.

$20=v_i+a(2.51)$

$20=v_i+2.51a$

Subtracting 2.51a from both sides, we get

$20-2.51a=v_i$ -----------equation(1)

Using another equation of motion

$v_f-v_i=2as$

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

$20-(20-2.51a)=2*a(82.90)$

Now, we need to solve it for a.

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So, the acceleration would be 0 m/s^2.

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2 years ago

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Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

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Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

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The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

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