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2 years ago

Guys PLZ I NEED HELP

Physics

2 answers:

BigorU [14]2 years ago

6 0

Answer:

its b sooo

Explanation:

klemol [59]2 years ago

4 0

Answer:

The answer is either D. (Hydrolysis) or C. (Abrasion)

Explanation:

I have taken the CST test for this.

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What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?

Sauron [17]

Explanation:

Given that,

Wavelength of light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Angle, $\theta=25^{\circ}$

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

$d\sin\theta=n\lambda$

n is 1 here

d is slit spacing

$d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m$

So, the slit spacing is $1.41\ \mu m$ .

6 0

3 years ago

Why do our eyes see the color red when we look at a tomato?

forsale [732]

Answer:

B. Tomatos reflect red light

Explanation:

The only reason colors exist is because the objects with color reflect all other light except for what they are portrayed as. White reflects all colors, and black absorbs all colors.

If you have any questions feel free to ask :)

4 0

3 years ago

Read 2 more answers

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

Given that water at standard pressure freezes at 0∘C, which corresponds to 32∘F, and that it boils at 100∘C, which corresponds t

barxatty [35]

Answer:

In two significant figure 360K

Explanation:

The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.

Hence,

ΔT = 212 - 32

ΔT = 180°F

To convert to °F to kelvin, we use the formula below

= (°F - 32) × 5/9 + 273.15

= (180°F - 32) × 5/9 + 273.15

= 355.37K ⇔ 360K

3 0

3 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers