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4 years ago

HELP PLZZZZZ Consider the following data:

Physics

2 answers:

Montano1993 [528]4 years ago

8 0

It’s A I’m sorry if it’s wrong

marishachu [46]4 years ago

4 0

Answer:

I think the answer is As you increase the current the resistance increases too.

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Ultrasound involves high-frequency sound waves. Sound waves can be used to image internal structures. This is the basis of

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Consider the diagram of a pendulum's motion shown above. A pendulum can be used to model the change from potential energy

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Explanation: Potential stores Kinetic Energy so, Point C will have Kinetic turned into Potential.

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3 years ago

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through diamond. The wavelength o

vladimir1956 [14]

Answer:

speed = 1.24 × 10⁸m/s

frequency = 4.74 × 10¹⁴Hz

wavelength = 262nm

Explanation:

the speed of the helium-neon light in zircon is given by,

v = c / n

c = 3 × 10⁸m/s is the speed of light in vacuum (and in air)

n = 2.419 is the refractive index of diamond

v = 3 × 10⁸ / 2.419

= 1.24 × 10⁸m/s

(b) Frequency

The wavelength of the light in air is:

λ₀ = 632.8 × 10⁻⁹

The frequency of the light does not depend on the medium, so it is equal in air and in diamond. Therefore, we can calculate the frequency by using the speed of light in air and the wavelength in air:

f₀ = c / λ₀

= 3 × 10⁸ / 632.8 × 10⁻⁹

= 4.74 × 10¹⁴Hz

and the frequency of the light indiamond is the same:

f¹ = f₀ = 4.74 × 10¹⁴Hz

To calculate the wavelength of the light in daimond, we can use the relationship between speed of light in diamond and frequency:

λ¹ = v / f¹

= 1.24 × 10⁸ / 4.74 × 10¹⁴

= 2.62 × 10⁻⁷m

= 262nm

3 0

3 years ago

In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is

tresset_1 [31]

Answer:

The expression for the initial speed of the fired projectile is:

$\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})$

And the initial speed ratio for the 9.0mm/44-caliber bullet is 1.773.

Explanation:

For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.

First Phase: Impact

In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:

$P_0=P_f\\v_0m=v_i(m+M)\\v_0=v_i\frac{m+M}{m}$ (1)

Second Phase: pendular movement

After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:

$Em_i=Em_f\\\frac{1}{2}mv^2_i=mgH\\v_i=[2gH]^\frac{1}{2}$ (2)

The height of the pendulum respect L and θ is:

$H=L(1-cos(\theta))$ (3)

Using equations (1),(2) and (3):

$\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})$ (4)

The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):

$\displaystyle \frac{v_{9mm}}{v_{44}} =\frac{(M+m_{9mm})m_{44}}{(M+m_{44})m_{9}}(\frac{1-cos(\theta_{9mm})}{1-cos(\theta_{44})} )^{\frac{1}{2}}=1.773$

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3 years ago

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Exercise to reduce stress

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4 years ago

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