All categories

3 years ago

What do extremely massive stars end their life as?

Physics

2 answers:

IrinaVladis [17]3 years ago

6 0

The most massivest stars end their lives as black holes. <em>(D)</em>

Andreyy893 years ago

3 0

I couldn’t find an exact but here is what I found

You might be interested in

A tennis ball is thrown into

liraira [26]

Answer:

Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.

Explanation:

7 0

2 years ago

Jack drops a stone from rest off of the top of a bridge that is 24.4 m above the ground. After the stone falls 6.6 m, Jill throw

alukav5142 [94]

-17.555m/s

first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.

then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.

then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2

4 0

3 years ago

A car advertisement states that a certain car can accelerate from rest to

kaheart [24]

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

$40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]$

We have to use the following kinematics equation:

$v_{f}= v_{i}+(a*t)$

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

6 0

3 years ago

Read 2 more answers

Longitudinal waves do not have

soldier1979 [14.2K]

E. Nonsense longitudinal waves have all of these properties

8 0

3 years ago

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride

marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

$\Delta x = 6.17 cm = 0.0617 m$

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

$2mg = k'\Delta x$ (1)

where

m = 76.2 kg is the mass of each children

$g=9.8 m/s^2$ is the acceleration of gravity

$k'$ is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

$k'=k+k=2k$

Substituting into (1) and solving for k, we find:

$2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m$

The period of the oscillating system is given by

$T=2\pi \sqrt{\frac{m}{k'}}$

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

$k'=2k=2(12,103)=24,206 N/m$ is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

$m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg$

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

In this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

$f=\frac{1}{2.09}=0.478 Hz$

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

$t=10T=10(2.09)=20.9 s$

#LearnwithBrainly

7 0

3 years ago