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3 years ago

Please help me with #6!

Physics

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

ok bet

Explanation:

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If the moon is a full moon tonight what will be the moon one week later waxing or waning

tatiyna

-- Starting from nothing (New Moon), the moon's shape grows ('waxes')
for half of the cycle, until it's full, and then it shrinks ('wanes') for the next
half of the cycle.

-- The moon's complete cycle of phases runs 29.53 days . . . roughly
four weeks.

-- So, beginning from New Moon, it spends about two weeks waxing until
it's full, and then another two weeks waning until it's all gone again.

-- After a Full Moon, the moon is waning for the next two weeks. So it's
definitely <em>waning</em> at <em><u>one week</u></em> after Full.

5 0

2 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago

Blake stands in a canoe in the middle of a lake. The canoe is stationary. Blake holds an anchor mass of 15 kg, then throws it we

Inessa05 [86]

The velocity of the canoe is 1.7 m/s.

<h3>What is momentum?</h3>

Momentum in physics is the products of mass and velocity. Now we have to find momentum with the formula; p = mv

a) Initial momentum = (15)8 m/s + 135 = 255 Kgms-1

b) Since momentum is conserved, the total momentum after throwing the anchor is still 255 Kgms-1

c) The final velocity of the boat is obtained from;

255 Kgms-1 = (15Kg + 135 Kg) v

v = 255 Kgms-1/(15Kg + 135 Kg)

v = 1.7 m/s

Learn more about momentum: brainly.com/question/904448

5 0

2 years ago

HELPPPP PLSSSS A group of students is investigating whether air resistance depends on the size of an object. The students throw

shutvik [7]

Answer:

Explanation: societal law is A statement about how things act in the natural world. the wire and electro magnet need to touch in question 2. a decrease in spee decreases kinetic energy. the test variable was the distance traveled by the scrunched paper.

8 0

2 years ago

Read 2 more answers

The red light from a helium-neon laser has a wavelength of 721.4 nm in air. Find the speed, wavelength, and frequency of helium-

saveliy_v [14]

Answer:

(a) the speed of helium-neon laser light in air is 3 x 10⁸ m/s

the wavelength of helium-neon laser light in air is 721.4 nm

the frequency of helium-neon laser light in air is 415.86 THz

(b) the speed of helium-neon laser light in water is 2.26 x 10⁸ m/s

the wavelength of helium-neon laser light in water is 542.4nm

the frequency of helium-neon laser light in water is 416.67THz

the wavelength of helium-neon laser light in glass is 480.9nm

the frequency of helium-neon laser light in glass is 415.88THz

From the results above, it can be seen that speed of the light is directly proportional to its wavelength, while the frequency of the light remained fairly constant for the different media.

Explanation:

Part (a) the speed, wavelength, and frequency of helium-neon laser light in air

Given;

wavelength of helium-neon laser light in air, λ = 721.4 nm

speed of light in air, v = 3 x 10⁸ m/s

v = f λ

where;

f is the frequency of helium-neon laser light in air

$f = \frac{v}{\lambda} = \frac{3*10^8}{721.4 *10^{-9}} =4.1586*10^{14} \ Hz$

f = 415.86 THz

Part (b) the speed, wavelength, and frequency of helium-neon laser light in water

refractive index of water = 1.33

$Refractive \ index \ of \ water =\frac{speed \ of \ light \ in \ air}{speed \ of \ light \ in \ water} = \frac{wavelength \ of \ light \ in \ air}{wavelength \ of \ light \ in \ water}$

$speed \ of \ light \ in \ water = \frac{speed \ of \ light \ in \ air}{Refractive \ index \ of \ water} \\\\speed \ of \ light \ in \ water = \frac{3*10^8}{1.33} = 2.26 *10^8 \ m/s$

Again;

$wavelength \ of \ light \ in \ water = \frac{wavelength \ of \ light \ in \ air}{Refractive \ index \ of \ water} \\\\wavelength \ of \ light \ in \ water = \frac{721.4 \ nm}{1.33} = 542.4 \ nm$

$f = \frac{v}{\lambda} = \frac{2.26*10^8}{542.4 *10^{-9}} =4.1667*10^{14} \ Hz$

f = 416.67 THz

Part (c) the speed, wavelength, and frequency of helium-neon laser light in glass

Refractive index of glass = 1.5

$speed \ of \ light \ in \ glass = \frac{speed \ of \ light \ in \ air}{Refractive \ index \ of \ glass} \\\\speed \ of \ light \ in \ glass = \frac{3*10^8}{1.5} = 2 *10^8 \ m/s$

Also;

$wavelength \ of \ light \ in \ glass = \frac{wavelength \ of \ light \ in \ air}{Refractive \ index \ of \ glass} \\\\wavelength \ of \ light \ in \ glass = \frac{721.4 \ nm }{1.5} = 480.9 \ nm$

$f = \frac{v}{\lambda} = \frac{2*10^8}{480.9 *10^{-9}} =4.1588*10^{14} \ Hz$

f = 415.88 THz

5 0

3 years ago

Read 2 more answers