What is the direction of acceleration due to gravity

light of a certain frequency has a wavelength of 438 nm in water.What is the wavelength of this light in benzene

sveticcg [70]

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

$\lambda (water) \times n(water) = \lambda (benzene) \times n(benzene)$

And, the wavelength of water is 438 nm

$\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]$

Now placing these values to the above formula

So,

$= 438 \times \frac{1.333}{1.501}$

= 388.97 nm

We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come

5 0

2 years ago

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

madreJ [45]

maximum static friction acting on the object will be

$F_s = \mu_s mg$

plug in all values

$F_s = 0.40 \times 15 \times 9.8 = 58.8 N$

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

6 0

2 years ago

An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w

Arturiano [62]

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

5 0

1 year ago

An object is launched with an initial velocity of 50.0 m/s at a launch angle of 36.9∘ above the horizontal. part a determine x-v

ankoles [38]

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

5 0

3 years ago

Read 2 more answers

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

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

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

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

$V_{x} = -1.40t + 22$