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3 years ago

What do you know about stars?

Physics

2 answers:

Gekata [30.6K]3 years ago

7 0

Blue stars are hotter than red stars

And the closest star to earth is our sun

Readme [11.4K]3 years ago

4 0

A star's life begins with the gravitational collapse of a gaseous nebula, when a stars mass is greater that .4 times our sun expands to become a red giant.

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A small object moves along the xx-axis with acceleration ax(t)ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At tt = 0 th

lora16 [44]

Answer:

x = 54.3m (on the +ve x axis)

Explanation:

This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.

Given that acceleration, a, is:

ax(t) = - 0.032(15.0 - t)

And the initial values are:

x(t = 0) = - 14.0m

v(t = 0) = 8.7m/s

Hence,

a = - 0.032(15 - t)

a = - 0.48 + 0.032t

a = dv/dt = -0.48 + 0.032t

To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:

v = ∫dv/dt = ∫(-0.48 + 0.032t)

∫dv = ∫(-0.48 + 0.032t)dt

(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)

Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:

=> v - 8.7 = -0.48t + 0.032t²/2

v = 8.7 - 0.48t + 0.016t²

To obtain distance, x, integrate the velocity and apply the initial values:

v = dx/dt = 8.7 - 0.48t + 0.016t²

=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)

∫dx= ∫(8.7 - 0.48t + 0.016t²)dt

(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)

Inputting the initial values t₀ = 0s, x₀ = - 14.0m:

(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3

x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0

Now that the distance, x, has been obtained, when t = 10s:

x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0

x = 87 - 24 + 5.3 - 14.0

x = 54.3m

Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).

5 0

3 years ago

Present at least one example that illustrates acceleration. Is this a scalar or vector quantity? Explain why. ...?

SIZIF [17.4K]

Acceleration is a vector because it has magnitude and direction

Example :
a Car stopping at the rate of 5 m/s. In this example, the direction of acceleration is the opposite of the direction of velocity, and the magnitude is 5 m/s

Hope this helps

5 0

3 years ago

Read 2 more answers

SP: Calculate the moment

ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, $F=80\ N$

Distance of application of force from the point about which moment is needed is, $d=25\ cm=\frac{25}{100}\ m=0.25\ m$

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

$M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}$

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0

3 years ago

a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b

SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0

3 years ago

The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

sukhopar [10]

Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

4 0

4 years ago