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MissTica
3 years ago
11

> Land Co

Physics
1 answer:
MrRissso [65]3 years ago
5 0

is the answer is the b

Explanation:

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The orbitals can be found by counting down the _____.
KengaRu [80]
I think it's B for electrons
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4 years ago
A 0.5kg bird has a PE of 137.2 J. What height is its perch?
jekas [21]

PE = mgh

137.2 = .5*9.8*h

h=137.2/(.5*9.8)

7 0
3 years ago
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When light goes from one material into another material having a HIGHER index of refractionA) its speed decreases but its wavele
Mnenie [13.5K]

Answer:

E) its speed and wavelength decrease, but its frequency stays the same

Explanation:

First of all, the frequency of a light wave does not depend on the medium, while wavelength and speed do. Therefore, the frequency remains costant.

In particular, the speed of light in a medium is given by:

v=\frac{c}{n}

where c is the speed of light in a vacuum and n is the index of refraction. From the formula, we see that v and n are inversely proportional: so, when the light moves into a material with higher index of refraction, its speed decreases.

Moreover, speed is related to wavelength by

v=\lambda f

where \lambda is the wavelength and f is the frequency. Since the two quantities are directly proportional, this means that since the speed decreases, the wavelength decreases as well.

So, the correct choice is

E) its speed and wavelength decrease, but its frequency stays the same

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3 years ago
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During the operation of a voltaic cell the cell produces
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Electrical energy spontaneously

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2 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
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