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:
brainly.com/question/18484687
Okay so yeah u have to minus then subtract then decide it it’s a method i was taught to do
Answer:
a) about 20.4 meters high
b) about 4.08 seconds
Explanation:
Part a)
To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.
In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:
Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:
Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)
To solve for "t" in this quadratic equation, we can factor it out as shown:
Therefore there are two possible solutions when each of the two factors equals zero:
1) t= 0 (which is not representative of our case) , and
2) the expression in parenthesis is zero:
Answer:
Newton's law of inertia - His first law states that every object remains at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... This is the first part cited in Newton's first law; "there is no net force on the airplane and it travels at a constant velocity in a straight line."
Newton's law of acceleration - "a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity."
Newton's law of Action/Reaction - "As a plane flies, the force of the air hitting the plane is always equal and opposite to the force of the plane pushing against the air. The force generated by the engine pushes against air while the air pushes back with an equal and opposite force."
Hope this helps! god bless :)
Please give me brainliest!
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad
