Answer:
1.75 m/s
Explanation:
k = Spring constant = 490 N/m
m = Mass of object = 5.8 kg
x = Displacement of spring = 0.19 m
v = Speed of object at the equilibrium position
The potential energy of the spring will balance the kinetic energy of the mass
![\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow v=\sqrt{\dfrac{490\times 0.19^2}{5.8}}\\\Rightarrow v=1.74637\ \text{m/s}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B%5Cdfrac%7Bkx%5E2%7D%7Bm%7D%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B%5Cdfrac%7B490%5Ctimes%200.19%5E2%7D%7B5.8%7D%7D%5C%5C%5CRightarrow%20v%3D1.74637%5C%20%5Ctext%7Bm%2Fs%7D)
The speed of the mass as it returns to the equilibrium position is 1.75 m/s.
PITCH
please mark brainliest!!!
Planets are not hot enough, and that's because they lack the internal energy sources that stars have: thermonuclear fusion (They do emit a little light, mainly infrared light, but nothing compared to the sun)
Planets don't have their own light, so simply they do not emit light, this phenomena goes back to their information. When they are formed, they are much, much less hotter than the stars. In stars, fusion reaction always goes on, in which energy, as well as light is formed, so, there is light for stars to emit for a long time.
Answer:
0.133 A
Explanation:
Given:
The voltage difference is, ![V=16\ V](https://tex.z-dn.net/?f=V%3D16%5C%20V)
The resistance of the circuit is, ![R=120\ ohms](https://tex.z-dn.net/?f=R%3D120%5C%20ohms)
Now, as per Ohm's Law, the potential difference across two points in a circuit is directly proportional to the current supplied in the circuit. The constant of proportionality is the resistance of the circuit.
Hence, the mathematical statement of Ohm's law is given as:
Where
is the current in the circuit.
Now, plug in 16 for 'V' and 120 for 'R' and solve for 'I'. This gives,
![16=120I\\I=\frac{16}{120}\\I=0.133\ A](https://tex.z-dn.net/?f=16%3D120I%5C%5CI%3D%5Cfrac%7B16%7D%7B120%7D%5C%5CI%3D0.133%5C%20A)
Therefore, the current in the circuit is 0.133 amperes.