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Zigmanuir [339]

2 years ago

12

If a sprinter’s mass is 60 kg, how much forward force must be exerted on the sprinter to make the sprinter accelerate at 0.8 m/s

2?

2 answers:

Natasha2012 [34]2 years ago

7 0

Forward force is 60×0.8
=48N

spin [16.1K]2 years ago

7 0

60 * 0.8 = 48
Thanks (:

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What is time and speed and acceleration

Andre45 [30]

Answer:

Acceleration is the rate of change of velocity with time.Acceleration occurs anytime an object's speed increases or decreases, or it changes direction. Much like velocity, there are two kinds of acceleration: average and instantaneous. Average acceleration is determined over a "long" time interval.

Explanation:

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2 years ago

Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo

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Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

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2 years ago

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

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

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

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

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$M =V \rho$

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V = Volume

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

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

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

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

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

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.

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mart [117]

It’s going to be both answer A and B but if you can only answer one then it’s going to be B

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Tems11 [23]

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HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

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3 years ago