To find black holes in space, scientists use telescopes to study visible light, X-rays, and radio waves. When gas orbits a black hole, it gets very hot and starts emitting X-rays and radio waves. Through special telescopes, the area surrounding the black hole looks really bright.
Answer: D)
Explanation:
In a RL circuit, as current can't change instantaneously, it starts from 0, till it reaches to the maximum possible value, according to Ohm's Law, i.e., E/R.
At any time, the current in the circuit (which is the same that passes through the inductor as it's a series circuit) is explained by the following equation:
I = E/R (1 - e-tR/L)
The quotient L/R is called the time constant of the circuit, and defines the time needed for the current reaches to its steady-state value.
If L is larger, the time constant will be larger, and it will take more time to the current to reach to its steady-state value.
Answer:
- it would probably snap or brake - that qustion makes no sense
Explanation:
Answer: 2.67 m/s
Explanation:
Given
Mass of block A is ![m_a=2\ kg](https://tex.z-dn.net/?f=m_a%3D2%5C%20kg)
mass of block B is ![m_b=3\ kg](https://tex.z-dn.net/?f=m_b%3D3%5C%20kg)
The initial velocity of block A ![u_a=5\ m/s](https://tex.z-dn.net/?f=u_a%3D5%5C%20m%2Fs)
the initial velocity of block B is ![u_b=0](https://tex.z-dn.net/?f=u_b%3D0)
After collision velocity of block A is ![v_a=1\ m/s](https://tex.z-dn.net/?f=v_a%3D1%5C%20m%2Fs)
Conserving momentum
![m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s](https://tex.z-dn.net/?f=m_au_a%2Bm_bu_b%3Dm_av_a%2Bm_bv_b%5C%5C%5C%5C2%5Ctimes%205%2B3%5Ctimes0%3D2%5Ctimes%201%2B3%5Ctimes%20v_b%5C%5C%5C%5Cv_b%3D%5Cdfrac%7B8%7D%7B3%7D%3D2.67%5C%20m%2Fs)
The momentum of block A after the collision is ![P_a=2\times 1=2\ kg.m/s](https://tex.z-dn.net/?f=P_a%3D2%5Ctimes%201%3D2%5C%20kg.m%2Fs)
Therefore, there is no change in sign.
Answer:
(D) the speed of the mass when the spring returns to its natural length is 1.58 m/s.
Explanation:
Given;
spring constant of the spring, k = 2 N/m
mass attached to the spring, m = 0.2 kg
compression of the spring, x = 0.5 m
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂m(v² - u²) = ¹/₂kx²
where;
u is the initial speed of the mass = 0
¹/₂mv² = ¹/₂kx²
mv² = kx²
Therefore, the speed of the mass when the spring returns to its natural length is 1.58 m/s.