Fahrenheit because the boiling point of water is 100 degrees Celsius which is 212 Fahrenheit which is very hot, and that would be about 200 Kelvin so therefore the answer is that the temperature was recorded in Fahrenheit not Kelvin or Celsius
Answer:
360 N
Explanation:
m = 30kg u = 2 m/s a = -2m/s/s
Since the object has an initial velocity of 2 m/s and acceleration of -2 m/s/s
the object will come to rest in 1 second but the force applied in that one second can be calculated by:
F = ma
F = 30 * -2
F = -60 N (the negative sign tells us that the force is acting downwards)
Now, calculating the force applied on the box due to gravity
letting g = -10m/s/s
F = ma
F = 30 * -10
F = -300 N (the negative sign tells us that the force is acting downwards)
Now, calculating the total downward force:
-300 + (-60) = -360 N
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<em>Hence, a downward force of 360 N is being applied on the box and since the box did not disconnect from the rope, the rope applied the same amount of force in the opposite direction</em>
Therefore tension on the force = <u>360 N</u>
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Answer:
25.6 m/s
Explanation:
Draw a free body diagram of the sled. There are two forces acting on the sled:
Normal force pushing perpendicular to the hill
Weight force pulling straight down
Take sum of the forces parallel to the hill:
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 38.0°)
a = 6.03 m/s²
Given:
v₀ = 0 m/s
a = 6.03 m/s²
t = 4.24 s
Find: v
v = at + v₀
v = (6.03 m/s²) (4.24 s) + (0 m/s)
v = 25.6 m/s
