Answer:
u are the very good person I know if u will do itself u will becam3 a rising star
Answer:
28.57 Mpc
Explanation:
This question is going to be solved by applying Hubble's Law.
This Hubble's Law is actually an observation in physical cosmology. This observation makes it clear that galaxies are moving away from the Earth, and are doing so at speeds proportional to their distance. This essentially means that the farther they are from the Earth, the faster they are moving away from Earth.
It is represented by this formula
v = H(0)D, where
v = speed
H(0) = Constant of proportionality, or otherwise, Hubble's constant.
D = Distance to a galaxy
Applying the given parameters to the formula, we have
v = H(0).D
D = v / H(0)
D = 2000 / 70
D = 28.57 Mpc
The particle has a constant horizontal velocity, and a vertical force won't affect the horizontal speed, so it should be fairly easy to find the last part, "the time taken for a 10m horizontal displacement," using a kinematic equation.
X = x + vt + (1/2)at²
10 = 0 + (1.6)t + (1/2)(0)t²
10/1.6 = t
t = 6.25s
So now we have to find the vertical displacement over 6.25 seconds on a particle of a 2.5kg mass with a force of 8N.
Start with Newton's second law.
F = ma
8 = (2.5)a
a = 3.2m/s²
Now, use kinematics again.
Y = y + vt + (1/2)at²
Y = 0 + (0)(6.25) + (1/2)(3.2)(6.25)²
Y = <u>62.5m</u>
Answer:
Explanation:
a )
Amplitude A = 14 mm , angular frequency ω = 2π / T
= 2π / .5
ω = 4π rad /s
φ₀ = initial phase
Putting the given values in the equation
14 = 25 cos(ωt + φ₀ )
14/25 = cosφ₀
φ₀ = 56 degree
x(t) = 25cos(4πt + 56° )
b )
maximum velocity = ω A
= 4π x 25
100 x 3.14 mm /s
= 314 mm /s
At x = 0 ( equilibrium position or middle point , this velocity is achieved. )
maximun acceleration = ω² A
= 16π² x A
= 16 x 3.14² x 25
= 3943.84 mm / s²
3.9 m / s²
It occurs at x = A or at extreme position.
Explanation:
Below is an attachment containing the solution.
