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2 years ago

Prisms seperate __ light, such as that from the Sun by wavelength

Physics

1 answer:

natka813 [3]2 years ago

3 0

Answer: white

Explanation:

prisms separate white light, such as that from the sun by wavelength.

Brainliest

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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.2 m. The mug s

ZanzabumX [31]

Part a can be solve using the equation of trajectory:

Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]

Where y is the height

X is the length

G is the acceleration due to gravity

Vo Is the initail velocity

a is the angle of trajectory

1.2 = 1.35 tan(0) + (9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]

Solve for V0 = 2.729 m/s

b. can be solve using the formula

v = sqrt(2gy)

= sqrt ( 2*1.2*9.81)

= 4.852 m/s going down ( 0 degree from the horizontal)

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3 years ago

What are the characteristics and phases of the moon

svetoff [14.1K]

Crescent, gibbous, waxing, and waning.

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3 years ago

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

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3 years ago

The part of the cerebrum that controls voluntary movements is located in the _______ lobe. A. frontal B. parietal C. temporal D.

uranmaximum [27]

The part of the cerebrum that controls Voluntary movements is located in the Frontal lobe.

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3 years ago

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Mr. Patel is looking for his cat in the garden at night. It is completely dark outside. Mr Patel shines his torch at the tree. W

UNO [17]

Answer:

the tree's light reflects into his eyes.

I think this is the ans

Explanation:

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2 years ago