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qwelly [4]
3 years ago
13

The brightest, hottest, and most massive stars are the brilliant blue stars designated as spectral class O. If a class O star wi

th a mass of 3.38 ´ 1031 kg has a kinetic energy of 1.30 ´ 1042 J, what is its speed?
Physics
1 answer:
aksik [14]3 years ago
4 0

Answer:

2.77 * 10^5 m/s

Explanation:

Let us recall that kinetic energy is given by 1/2 mv^2

Where;

m = mass of the body

v = velocity of the body

In this case,

m = 3.38 * 10^31 kg

KE= 1.30 * 10^42 J

KE = 1/2 mv^2

v = √2KE/m

v = √2 * 1.30 * 10^42/3.38 * 10^31

v = √7.69 * 10^10

v = 2.77 * 10^5 m/s

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#ClutchMoments lol !
4 0
3 years ago
. A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that
ZanzabumX [31]

Answer:

\theta=32rad

Explanation:

In an uniformly accelerated circular motion, the angle traveled by the object is given by:

\theta=\frac{\omega_f+\omega_0}{2}t

Here \omega_f is the final angular speed, \omega_0 is the initial angular speed and t is the time of the motion. Replacing the given values:

\theta=\frac{12\frac{rad}{s}+4\frac{rad}{s}}{2}(4s)\\\theta=32rad

5 0
3 years ago
3 a A motorcyclist starts from rest and reaches
andrew-mc [135]

The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.

<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time taken

v - u/t = a

a = 6 - 0/3

= 2 m/s^2

Again;

v - u/t = a

a = 0 - 6/5

a = - 1.2m/s^2

Learn more about acceleration:brainly.com/question/12550364

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7 0
1 year ago
Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
Electro negativity increases when atoms __. (Apex)
abruzzese [7]
C. I took the test...........
3 0
3 years ago
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