Answer:
388.97 nm
Explanation:
The computation of the wavelength of this light in benzene is shown below:
As we know that
n (water) = 1.333
n (benzene) = 1.501

And, the wavelength of water is 438 nm
![\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]](https://tex.z-dn.net/?f=%5Clambda%20%28benzene%29%20%3D%20%5Clambda%20%28water%29%20%5B%5Cfrac%7Bn%28water%29%7D%7Bn%28benzene%7D%5D)
Now placing these values to the above formula
So,

= 388.97 nm
We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come
maximum static friction acting on the object will be

plug in all values

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force
So here it is given that applied force is 20 N
so here object will not move due to this force and it will remain at rest always
due to this applied force
The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.
<h3>
UF cable</h3>
UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.
For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.
Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.
Learn more about UF cable here: brainly.com/question/8591560
Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)

a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s


The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
![[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%2822%20-%2011.78%29%20%2A%20%287.30%29%5D%20%20%2B%20%5B%2811.78%20-%200%29%20%2A%20%287.30%29%5D)
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m