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2 years ago

6

find the following Distance, Time, and Velocity in the problem; a track runs a 100-m dash in 11 seconds. What is his average vel

ocity during the run? (please include explanation and formula thankyou)

1 answer:

kow [346]2 years ago

5 0

Answer:

9.09m/s

Explanation:

now velocity of a body is measure of the distance covered by a body in a particular time frame,so simply its the rate at which distance is covered,

thus mathematically V(velocity)=d(distance covered)/t(time it takes to cover the distance)

hence V=100m/11s

which is V=9.09m/s.

so this means the runner covers 9.09m in every second ,<em> </em><em>well</em><em> </em><em>not</em><em> </em><em>bad</em><em> </em><em>lol</em><em> </em><em>!</em>

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A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc

vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

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3 years ago

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3 years ago

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

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3 years ago

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ki77a [65]

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3 years ago

a motorcycle starts from rest covers 200 meter distance in 6 second calculate final velocity and acceleration

Ivanshal [37]

Explanation:

s = ut + 1/2 a t^2

200 = 0 * 6 + 1/2 * a * (6)^2

200 = 1/2 * a * 36

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a = 200/18

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v = u + at

v = 0 + 11.1 * 6

v = 66.6m/s

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3 0

2 years ago