This is true. Gravity is constantly pulling on anything and everything (even light!), no matter how far away it is from another object.
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
![[m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a](https://tex.z-dn.net/?f=%5Bm_1v_1%2Bm_2v_2%5D_b%3D%5B%28m_1%2Bm_2%29v%5D_a)
Filling in:
![[1200(4.5)+2100(0)]=[(1200+2100)v]](https://tex.z-dn.net/?f=%5B1200%284.5%29%2B2100%280%29%5D%3D%5B%281200%2B2100%29v%5D)
which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
The only correct statement on the list is choice-A./
Answer:
7.6 g
Explanation:
"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.
The heat gained by the copper, water, and ice = the heat lost by the steam
Heat gained by the copper:
q = mCΔT
q = (120 g) (0.40 J/g/K) (40°C − 0°C)
q = 1920 J
Heat gained by the water:
q = mCΔT
q = (70 g) (4.2 J/g/K) (40°C − 0°C)
q = 11760 J
Heat gained by the ice:
q = mL + mCΔT
q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)
q = 4880 J
Heat lost by the steam:
q = mL + mCΔT
q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)
q = 2452 J/g m
Plugging the values into the equation:
1920 J + 11760 J + 4880 J = 2452 J/g m
18560 J = 2452 J/g m
m = 7.6 g
to water at