Answer:
2200000 = 2.2E6 min for light from Proxima to reach earth
8.3 min from light sun to reach earth
2.2E6/8.3 = 2.56E5 times for light from Proxima
Proxima is about 256,000 times farther away than the sun
Since the sun is about 93,000,000 = 9.3E7 miles from earth
Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away
Note - the speed of light is
3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given
Answer: a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Explanation:
Acceleration is the rate of change in the velocity per time
a = change in velocity/time
a = ∆v/t
average acceleration a = (v2 -v1)/t. ....1
Given;
Final velocity v2 = 1.63m/s
Initial velocity v1 = -1.15ms
time taken t = 2.11s
Substituting into eqn 1
a = [1.63 - (-1.15)]/2.11
a = (1.63+1.15)/2.11
a = 2.78/2.11
a = 1.32m/s2
Therefore, the average acceleration is 1.32m/s2
Answer:
According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."
Explanation:
Hope this helps! :)
Answer: A projectile which is fired horizontally is being constantly acted upon by acceleration due to gravity, acting vertically downwards. Hence, it does not follow a straight line path. Also Why a projectile fixed along the horizontal not follow a straight line path? Because the projectile fired horizontally is constantly acts upon by acceleration due to gravity acting vertically downwards.
Explanation:
Hope this helped :)
Explanation:
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass
as


Substituting (2) into (1), we get

where
, the frictional force on
Set this aside for now and let's look at the forces on 
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>
Let the x-axis be (+) up along the inclined plane. We can write the forces on
as


From (5), we can solve for <em>N</em> as

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

Substituting (7) into (4) we get

Collecting similar terms together, we get

or
![a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)](https://tex.z-dn.net/?f=a%20%3D%20%5Cleft%5B%20%5Cdfrac%7Bm_B%5Csin30%20-%20%5Cmu_km_A%7D%7B%28m_A%20%2B%20m_B%29%7D%20%5Cright%5Dg%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%288%29)
Putting in the numbers, we find that
. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get
. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get 