Fossil fuels like coal, natural gas and oil.
Hopes this helps!
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 
Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:

n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:

B) For the 3rd harmonic you have:

C) For the 2nd harmonic:

Answer:
Basic kinematics, negating drag and assuming ideal conditions, we use the equation:
d=vi*t+1/2*a*t^2
Since vi is 0 (we know this because you’re dropping it, not throwing it)…
…and the only acceleration acting on it is gravity, a=9.8 m/s^2…
…we get
d=1/2(9.8)(5)^2
Explanation:
Some quick mental math tells us that this is about 125 m.
Plugging it in, we find it to be 122.5 m.
Answer:
1902.75 kg
Explanation:
From Law of conservation of momentum,
m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1
make m₂ the subject of the equation,
m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2
Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity
Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)
Substituting into equation 2.
m₂ =[2537(8) - 2537(14)]/(0-8)
m₂ = (20296-35518)/-8
m₂ = -15222/-8
m₂ = 1902.75 kg.
Thus the mass of the car = 1902.75 kg