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3 years ago

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Arrange the following in order of their appearance in the life cycle of a star: white dwarf, red giant, main-sequence star. Expl

ain your answer.

1 answer:

Goryan [66]3 years ago

6 0

Main-sequence star, red giant, and white dwarf

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a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i

Katyanochek1 [597]

Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>

Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration

Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

<em>ANSWER: 240 N </em>

Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²

Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m

<em>ANSWER: d = 13.5 m</em>

3 0

3 years ago

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

Scilla [17]

Answer:

d= 5.62 m

Explanation:

In order to have destructive interference, the path difference from the sources to the listener must be an odd multiple of half wavelengths, as follows:

d = (2n+1) * λ/2

In orfer to know which is the wavelength, we can use the relationship between propagation speed (in this case speed of sound), frequency and wavelength:

v= λ*f ⇒ λ = v/f = 344 m/s / 688 1/sec = 0.5 m

So, the path difference must be, at least, λ/2:

d = b-a = λ/2, where b is the distance to the speaker B, and a, the distance to the speaker A.

Applying Pithagorean Theorem, as the perpendicular distance d (which is our unknown) is the same for the triangles defined by the horizontal distance to the listener, and the straight line from the new position to the sources, we can write:

d² = a²- (3.0)²

d² = b²- (3.5)²

As the left sides are equal, so do right sides:

a² - (3.0)² = b² - (3.5)²

⇒ b² - a² = (3.5)² - (3.0)² = 3.25

We can replace (b²- a²) as follows:

b² - a² = (b+a)(b-a) = 3.25 (2)

We know that b-a = λ/2 = 0.25

Replacing in (2), we have:

b+a = 3.25 / 0.25 = 13 m

b-a = 0.25 m

Adding both sides:

2*b = 13.25 m ⇒ b= 13.25 /2 = 6.63 m

⇒ d² = (6.63)² - (3.5)² = 31.6 m²

⇒ d=√31.6 m² = 5.62 m

3 0

3 years ago

What is the component of d1 along the direction of d2?

aniked [119]

I don't know lol hahahahhahahahahahahahahahahahahaha

8 0

3 years ago

What height will the object reach? 12 points. Will give brainliest.

umka21 [38]

Answer:

12.7 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 56.7 Km/hr

Maximum height (h) =..?

First, we shall convert 56.7 Km/hr to m/s. This can be obtained as follow:

Initial velocity (m/s) = 56.7 x 1000/3600

Initial velocity (m/s) = 15.75 m/s

Next, we shall determine the time taken to get to the maximum height. This can be obtained as follow:

Initial velocity (u) = 15.75 m/s

Final velocity (v) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

v = u – gt (since the ball is going against gravity)

0 = 15.75 – 9.8 × t

Rearrange

9.8 × t = 15.75

Divide both side by 9.8

t = 15.75/9.8

t = 1.61 secs.

Finally, we shall determine the maximum height as follow

h = ½gt²

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 1.61 secs.

Height (h) =..?

h = ½gt²

h = ½ × 9.8 × 1.61²

h = 4.9 x 1.61²

h = 12.7 m

Therefore, the maximum height reached by the ball is 12.7 m

3 0

3 years ago

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

nata0808 [166]

<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Now in the given problem

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

u = 51.5 m/s, for maximum range θ = 45⁰

So maximum distance reached = $\frac{51.5^2sin(2*45)}{9.81}=270.36 meter$

So it will go a maximum of 270.36 meter.

5 0

3 years ago