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Neporo4naja [7]
3 years ago
6

Which of the following statements about horizons is true?. a. They are characterized by whether they are solid, liquid, or gas..

b. They are represented by Roman numerals.. c. All soils have completely different horizon patterns.. d. They are defined by their different physical features.
Physics
1 answer:
garik1379 [7]3 years ago
3 0
<span>The answer is d. they are defined by their different physical features. A soil horizon is a layer in a soil profile. Horizons are defined by different physical features such as colour or texture. Other features include, structure and permeability. Each horizon is represented by a letter; A,B,C....</span>
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A charge of 8.4*10^-4 moves at an angle of 35 degrees to a magnetic field that has a field strentgh of 6.7
12345 [234]

The force on a charged particle in a magnetic field is given by

the speed of the charged particle = 10842 m/s.

Explanation:

F= q V B sinθ

F=force=3.5 x 10⁻²N

q= charge= 8.4 x 10⁻⁴ C

B= magnetic field= 6.7 x 10⁻³ T

θ=35⁰

Thus the velocity is given by V=\frac{F}{q B sin35}

V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]

V=10842 m/s

3 0
3 years ago
HELP!!! I have no idea how to calculate this.
stiks02 [169]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the kinematics of a body.

Since, we know that Velocity = Distance / time

hence, V = 20/5 = 4 m/s

hence the velocity of the RC car is 4 m/s westwards direction.

4 0
3 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

4 0
3 years ago
Your friend wants to join the school track team, and has asked for your help to determine how fast she can run. What kind of inf
Zigmanuir [339]

Answer:

you calculate a specific type of run for example 100m and it takes 20 seconds to finish and calculate the time it takes them to finish

hope this helps

have a good day :)

Explanation:

6 0
3 years ago
A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

  • u = 8m/ s
  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

7 0
3 years ago
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