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NeX [460]
3 years ago
12

Compared to a blue star in the same local cluster, the surface temperature of a red star is a) greater. b) the same. c) lower. d

) not consistently any of these
Physics
1 answer:
Andrew [12]3 years ago
3 0

Hello! My name is Zalgo and I am here to help you out on this concluding day. The answer would be C);lower. The reason it would be lower is because the hottest color of flames would be blue. Considering the way a start emits light is fire, this would be the most logical reason for it.

I hope that this helps! :P

"Stay Brainly and stay proud!" - Zalgo

(By the way, do you mind marking me as Brainliest? I'd greatly appreciate it! Thanks! X3)

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For her birthday, amy received flowers that had a dull red appearance. compared with the entire range of visible light waves, th
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The flowers reflect relatively low frequency and low amplitude light waves.
White light contains a spectrum of colors, and red light is at the beginning of this spectrum in terms of energy. This means that it has the lowest frequency, as the energy of a wave of light is directly proportional to its frequency.
The amplitude of a light wave gives a measure of how intense the light is. A dull light means that the amplitude of the wave is low.
4 0
3 years ago
What is potential energy?
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energy associate with position or shape
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3 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
3 years ago
A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.​
Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0
3 years ago
Which layer of the atmosphere is the top layer of the thermosphere?
joja [24]
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8 0
3 years ago
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