Find the acceleration of a box of mass 3.0 kg, if applied force is 20.0 N at 37° above the x axis, and the coefficient of kineti
1 answer:
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Answer:
B) t = 1.83 [s]
A) y = 16.51 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity = 0
Vo = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81 [m/s²]
t = time [s]
Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.
A) The maximum height is reached when the final velocity of the ball is zero.
0 = 18 - (9.81*t)
9.81*t = 18
t = 18/9.81
t = 1.83 [s], we found the answer for B.
Now using the following equation.
where:
y = elevation [m]
Yo = initial elevation = 0
y = 18*(1.83) - 0.5*9.81*(1.83)²
y = 16.51 [m]
To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 = <em>6</em><em>.65 μm</em>
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 = <em>6</em><em>.65 μm</em>