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9966 [12]
3 years ago
5

The blackbody curve for a star named Zeta is shown below. What is the peak wavelength for this star?

Physics
2 answers:
VladimirAG [237]3 years ago
8 0

860 nanometers.

Hope this helps.

hram777 [196]3 years ago
3 0

Answer:

The answer is 860 nm!

Just took the test

Explanation:

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What is kinetic energy
mixas84 [53]
<span>(symbol K)</span><span> Energy that an object possesses because it is in motion. It is the energy given to an object to set it in motion; it depends on the mass (</span>m) of the object and its velocity (v<span>), according to the equation K = 1/2 </span>mv2<span>. On impact, it is converted into other forms of energy such as heat, sound and light.</span>
8 0
3 years ago
A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel
mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

\vec v_{ra} = 12 m/s North

so it will be

Let North direction is along Y axis and East direction is along X axis

\vec v_{ra} = 12\hat j

also it is given that speed of air is 6.7 m/s relative to ground

\vec v_a = 6.7 \hat i

now as we know by the concept of relative motion

\vec v_{ab} = \vec v_a - \vec v_b

\vec v_{ra} = \vec v_r  - \vec v_a

now by rearranging the terms

\vec v_r = \vec v_{ra} + \vec v_a

\vec v_r = 12 \hat j + 6.7 \hat i

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

v_r = \sqrt{12^2 + 6.7^2}

v_r = 13.7 m/s

so the net speed of Robin with respect to ground will be 13.7 m/s

7 0
3 years ago
If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose
Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it  has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency  f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0
3 years ago
Can anyone fill in the blanks for the potential and kentic energy? Also, is this showing energy transformation? Thank you so muc
Bezzdna [24]

Potential Kinetic Kinetic Potential Potential


4 0
3 years ago
of the following and draw a neat, properly labelled free body diagram. Indicate the size of forces with the lengths of the arrow
julsineya [31]

We will have that the diagram will be the following:

Here "F" is the force the shuttle is using to take off, friction is the friction with the air and "W" is the weight of the shuttle.

4 0
1 year ago
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