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3 years ago

Compare tidal range to spring tide

1 answer:

7 0

Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.

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Are There some sounds for which we are all deaf

lana66690 [7]

Answer: Infrasound is the span of low-frequency sounds below 20 Hz that fall below the hearing range of humans. While these sounds escape our ears, scientific instruments can detect them—and tell us some interesting things about the planet.

Explanation:

6 0

3 years ago

A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s.

Bogdan [553]

Answer:

The time of Mars is 1.65 times larger on Mars than on Earth

Explanation:

The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole

Vf = Vo - V pothole

B) let's transform the weight of free groin system and N international system

1 N = 0.2248 lb

2.8 lbs (1N / 0.2248lbs) = 12.5 N

c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity

Y = Vo t - ½ g t²

Y = - ½ g t²

t = √ 2Y / g

Mars

gm = 0.37g

gm = 0.37 9.8

gm = 3,626 m / s²

t = √( 2 1.9 / 3.626 )

t = 1.02 s

Earth

t = √( 2 1.9 / 9.8)

t = 0.62 s

To make the comparison of time we are the relationship between the two

tm / te = 1.02 / 0.62

tm / te = 1.65

The time of Mars is 1.65 times larger on Mars than on Earth

7 0

2 years ago

Read 2 more answers

NEED HELP NOW!!! I WILL BRAINLIEST.

Sergio039 [100]

Answer:

1. Carbon-12

2. Nitrogen-13

3. Carbon-13

7 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

distance travel in the reaction time

d₁ = v x t

d₁ = 20 x 0.5 = 10 m

distance travel after you apply brake

using equation of motion

v² = u² + 2 a s

0 = 20² - 2 x 10 x s

s = 20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

distance between car and the deer = 38 m - 30 m

= 8 m

b) now, maximum speed car.

distance travel in reaction time

d₁ = s x t

d₁ = 0.5 V

distance left between them

d₂ = 38 - d₁

d₂ = 38 - 0.5 V

distance travel after you apply brake

using equation of motion

v² = u² + 2 a d₂

0 = (V)² - 2 x 10 x (38 - 0.5 V)

V² + 10 V - 760 = 0

now, solving the quadratic equation

$x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}$

V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

7 0

3 years ago