The bicyclist accelerates with magnitude <em>a</em> such that
25.0 m = 1/2 <em>a</em> (4.90 s)²
Solve for <em>a</em> :
<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²
Then her final speed is <em>v</em> such that
<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)
Solve for <em>v</em> :
<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s
Convert to mph. If you know that 1 m ≈ 3.28 ft, then
(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h
The most common one is junk food and sugary snacks are very very addicting! they are almost like drugs!!!
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
