The distance of the galaxy is 32.86 Mpc.
Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.
Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have
D = v/H₀
Substituting the values of the variables into the equation, we have
D = 2300 km/s ÷ 70 km/s/Mpc
D = 32.86 Mpc
So, the distance of the galaxy is 32.86 Mpc
Learn more about hubble law here:
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Answer:(10.69, 11.436)
Explanation:
Given
initial height of ball is 2 m
height of basket is 3.05 m
Launching angle

y=1.05
equation of trajectory of ball is given by

for x=12.27

u=10.69
for x=11.73

u=11.436 m/s
Thus range of speed is (10.69, 11.436)
Answer:
gravitational force
electrostatic force
Explanation:
The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.
The Mutual force of gravitational pull that they exert on each other can be given as:

where:
gravitational constant 
are the masses of individual balloons
the radial distance between the center of masses of the balloons.
But when there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.
Given as:

where:

are the charges on the individual balloons
R = radial distance between the charges.
From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.
We need to find its height.
We will use the formula P.E = mgh
Therefore h = P.E / mg
where P.E is the potential energy,
m is mass in kg,
g is acceleration due to gravity (9.8 m/s²)
h is the height of the object's displacement in meters.
h = P.E. / mg
h = 14000 / 40 × 9.8
h = 14000 / 392
h = 35.7
Therefore the canon ball was 35.7 meters high.
I this is tricky one sec.. I would go with b