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horsena [70]
4 years ago
6

The star Betelgeuse has a large negative absolute magnitude and a large positive apparent magnitude. What does this mean? A. Bet

elgeuse appears to be very bright, but actually is very dim. B. Betelgeuse appears to be very dim, but is actually very bright. C. Betelgeuse appears to be close to us, but is actually far away.
Physics
1 answer:
Papessa [141]4 years ago
5 0
I think it is "A" hope it help
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An ice skater has a moment of inertia of 2.5 kg.m2 when her arms are extended and a moment of inertia of 1.2 kg.m2 when her arms
Damm [24]

Answer:

\omega=4.7916\ rad.s^{-1}

Explanation:

Given:

  • moment of inertia of the skater with extended arms, I'=2.5\ kg.m^2
  • moment of inertia of the the skater with pulled-in arms, I=1.2\ kg.m^2
  • angular velocity of the skater with extended arms, \omega'=2.3\ rad.s^{-1}

<u>Using the law of conservation of angular momentum:</u>

I'.\omega'=I.\omega

2.5\times 2.3=1.2\times \omega

\omega=4.7916\ rad.s^{-1}

6 0
3 years ago
Which part of the electromagnetic specturm has the lowest radiant energy?
atroni [7]
Radio waves have the lowest radiant energy, hope this helps :]
4 0
3 years ago
If Angela ran to the bus and back to where she started what distance would she travel ?
Yakvenalex [24]

Answer:

Distance = 30 m

Explanation:

Since the position of the bus is at zero, and distance is a scalar quantity. That is, we are only concerned about the magnitude

If Angela ran to the bus and back to where she started, the distance travelled will be:

Distance = 15 + 15 = 30 m

But her displacement will be:

Displacement = 15 - 15 = 0.

5 0
3 years ago
The thing that is circled, correct me if im wrong please.
Irina18 [472]
It looks correct, but that graph only shows position and time?
5 0
3 years ago
Read 2 more answers
A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t
Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height  is given by;

v = u - gt

u = gt

t = \frac{u}{g}

Time of flight, T = 2t

T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0
3 years ago
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