Answer: 27 joules
Explanation:
Work is done when force is applied on the bench over a distance. it is measured in joules.
Workdone = force x distance
= 45 N x 0.6 metres
= 27 joules
Thus, 27 joules of work is done on the bench.
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
Can you please stop pasting this question, just go to his profile and ask him.
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
