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2 years ago

What is the acceleration of a racing car if it’s speed is increased uniformly form 44m/s to 66 m/s over an 11 sec. period.

Physics

1 answer:

jonny [76]2 years ago

7 0

Answer:

$a=2\ m/s^2$

Explanation:

Given that,

Initial velocity, u = 44 m/s

Final velocity, v = 66 m/s

Time, t = 11 s

Acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\a=\dfrac{66-44}{11}\\\\a=2\ m/s^2$

So, the acceleration of the car is $2\ m/s^2$ .

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Answer:

D) Since the stars would move from East to West just as the Sun and Moon do.

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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

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

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

Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.

valentina_108 [34]

The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics

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3 years ago

The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.What power is absorbed by the bulb?

Verizon [17]

Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

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Given:

Voltage = 3 V

Current = 200 mA = 0.200 A

Conversion used : (1 mA = 0.001 A)

Now put all the given values in the above formula, we get:

Power = Voltage × Current

Power = 3V × 0.200 A

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Thus, the power absorbed by the bulb is, 0.600 W

3 0

3 years ago

Mountains are part of which earth sphere?? Am I correct?

sammy [17]

You are correct. Mountains are part of the lithosphere.

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3 years ago

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