Ask question

Ask question

All categories

3 years ago

10

A truck travels up a hill with a 7.5◦incline.The truck has a constant speed of 24 m/s.What is the horizontal component of thetru

ck’s velocity?Answer in units of m/s.005 (part 2 of 2) 10.0 pointsWhat is the vertical component of the truck’svelocity?Answer in units of m/s.

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Answer:

The horizontal component of the truck's velocity is: 23.70 m/s

The vertical component of the truck's velocity is: 3.13 m/s

Explanation:

You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.

The identities are:

Cosα= $\frac{CA}{H}$

Senα= $\frac{CO}{H}$

Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus

The horizontal component of the truck's velocity is:

Let Vx represent it.

In this case, CA=Vx, H=24 and α=7.5 degrees

Vx=(24)Cos(7.5)

Vx=23.79 m/s

The vertical component of the truck's velocity is:

Let Vy represent it.

In this case, CO=Vy, H=24 and α=7.5 degrees

Vy=(24)Sen(7.5)

Vy=3.13 m/s

You might be interested in

What is a wave period?

Kamila [148]

A wave period is the time it takes to complete one cycle

I hope that's help:0

8 0

3 years ago

Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the

polet [3.4K]

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

$\rm \theta$ is the angle formed between the path of the incident light and the diffracted light = 9. 2°

λ is the wavelength of the light=490nm=4.9

N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

$n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\ d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm$

The number of lines per mm is found as;

$\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm$

Hence the number of lines per mm in the diffraction grating is 326.

To learn more about diffraction grating refer to the link;

brainly.com/question/1812927

5 0

2 years ago

Which phrase most closely characterizes a person's sense of self?

vampirchik [111]

My guess would be choice D

7 0

3 years ago

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

3 years ago

The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

5 0

3 years ago

Read 2 more answers