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2 years ago

How does proximity to oceans affect Temperatures and rainfall amounts?

Physics

1 answer:

vampirchik [111]2 years ago

4 0

Answer:

Because they absorb and emit heat more steadily than land, oceans help moderate temperatures in coastal areas, making winters warmer and summers cooler. Winds that blow in from across the oceans bring more rainfall and higher humidity.

Explanation:

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Because weight is a measurement of the amount of gravity pulling on an object, weight is considered a ______. Question 18 option

Andrews [41]

Answer: Force I believe.

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1 year ago

Which statement describes an example of static electricity?

lana [24]

Answer: Negatively charged particles are repelled by other negatively charged particles

Explanation:

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3 years ago

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A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initi

Gelneren [198K]

Mass of gold m₁ = 47 g

Initial temperature of gold T₁ = 99 C

Specific heat of gold C₁ = 0.129 J/gC

final temperature T₂ = 38 C

Heat needed by the gold to cool down

Q =m₁ * C₁* ( T₁ - T₂)

Q = (47)(0.129)(99-38)

Q = 369.843 J

This heat will be given by the water

we need to find out mass of water m₂

and initial temperature of water is T₃ = 25 C

Specific heat of water C₂ = 4.184 J/gC

Q = m₂*C₂*(T₂ - T₃)

369.843 = m₂(4.184)(38-25)

m₂ = 6.8 g

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2 years ago

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You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

$f(x,t)=Acos(kx-\omega t)$

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

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2 years ago

A car with mass 1500 kg moves with constant velocity of 36 m/s. The driver sees a group of cows in front and he immediately step

Crazy boy [7]

From laws of motion:

$S = ( \frac{v + u}{2} ) \times t$
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s

Substitute the values, hence:
$S = ( \frac{0 + 36}{2} ) \times 6$
$S = (18) \times 6 \\ \\ S = 108m$

But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.

Therefore, the distance between the car and the cows = 160-108
Distance = 52m

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3 years ago