Answer:<u><em>B. Temperature</em></u>
Explanation:The temperature of the star such as Sun is measured. Using this measurement, its peak wavelength and energy can be determined.For determination of wavelength, <u>Wien's displacement law is used</u>. This law states that, the sun like body emits all kinds of wavelengths and thus is nearly a black body.For black body, the peak wavelength emitted is inversely proportional to the temperature of the body. From the wavelength, energy can be calculated.<u>Temperature is the property which is primarily responsible for determining the type of electromagnetic energy and peak wavelength emitted by star</u>.
Answer:
B. 0.16 m
Explanation:
The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.
Since the dart is thrown horizontally, the initial vertical velocity is zero:

While the horizontal velocity is

The horizontal distance covered is

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

Answer:
I=P/U=6/12=0.5(A)
Explanation: P=UI ( CÔNG SUẤT = HIỆU ĐIỆN THẾ NHÂN VỚI C Đ D Đ)
Answer:
a) t = 3.35[s]; b) t = 1.386[s]
Explanation:
We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].
We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.
And using kinematics equations:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%3D0%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Cy%3D55%20%5Bm%5D%5C%5Cv_%7Bf%7D%5E%7B2%7D%3D0%2B2%2A9.81%2A55%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B2%2A9.81%2A55%7D%20%5C%5Cv_%7Bf%7D%3D32.85%5Bm%2Fs%5D)
Now we can calculate the time:
![v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Bg%2At%5C%5Ct%3D%5Cfrac%7Bv_%7Bf%7D-v_%7Bo%7D%7D%7Bg%7D%5C%5C%20t%3D%5Cfrac%7B32.85-0%7D%7B9.81%7D%5C%5C%20t%3D3.35%5Bs%5D)
Now we can calculate the second time, but using as a initial velocity 32.85[m/s].
The final velocity will be:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cv_%7Bf%7D%3D%5Csqrt%7Bv_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%7D%20%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B32.85%5E%7B2%7D%2B2%2A9.81%2A55%20%7D%20%5C%5Cv_%7Bf%7D%3D46.45%5Bm%2Fs%5D)
Now we can calculate the second time:
![t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B46.45-32.85%7D%7B9.81%7D%20%5C%5Ct%3D%201.386%5Bs%5D)
Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.