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3 years ago

Tides are a result of

Physics

2 answers:

Virty [35]3 years ago

6 0

The answer should be D but if its not then im sorrry, idk

Mrrafil [7]3 years ago

6 0

Tides are a result of <u>"the gravitational attraction of the Sun and Moon".</u>

An especially high tide (Spring tide) happens when the sun and moon are in line and both pull in a similar way.

Tides may appear to be straightforward at first glance, yet the intricate details of tides jumbled extraordinary logical masterminds for a considerable length of time they even driven Galileo astray into a bunk theory.

Today individuals realize that the gravitational pulls between the earth, moon and sun manage the tides. The moon, however, impacts tides the most.

The moon's gravitational pull on the earth is solid enough to pull the seas into bulge. On the off chance that no different powers were impacting everything, shores would encounter one high tide a day as the earth pivoted on its hub and coasts kept running into the seas' bulge confronting the moon.

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A child is pushing a merry-go-round. the angle through which the merry-go-round has turned varies with time according to θ(t)=γt

Andreyy89

I attached the missing part of the question.
Part A
Angular velocity is simply the rate at which angle changes, in other words, it is the first derivative of the angle function with respect to time:
$\theta (t)=\gamma t+\beta t^3\\ w(t)=\frac{d\theta(t)}{dt}\\ w(t)=\gamma +3\beta t^2$
Part B
Initial value is the value at t=0. To find initial value we simply plug t=0 into the equation.
$w(0)=\gamma +3\beta\cdot 0^2\\
w(0)=\gamma\\
w_0=\gamma= 0.422\frac{rad}{s}$
Part C
To find these values we simply need to plug in t=5 in the equation.
$w(5)=\gamma +3\beta\cdot 5^2\\
w(5)=\gamma +75\cdot \beta=0.422+75 \cdot 1.35\cdot10^{-2}=1.43\frac{rad}{s}$
Part D
Average angular velocity is total angular displacement divided by time:
$w_{av}=\frac{\Delta \theta}{\Delta t}$
$\Delta \theta=\theta(5)-\theta(0)=\gamma\cdot5+\beta 5^3\\
\Delta \theta=0.422\cdot5+125\cdot1.35\cdot10^{-2}=3.7975 $rad$
The average angular velocity is:
$w_{av}=\frac{\Delta \theta}{\Delta t}=\frac{3.7975 }{5}=0.7595\frac{rad}{s}$

3 0

3 years ago

he force of gravity is an inverse square law. This means that, if you double the distance between two large masses, the gravitat

Kryger [21]

The force decreases by a factor of 4, because 4 is 2-squared. (C).

And they don't have to be "large" masses. The force behaves exactly the same between the lint in your pocket and a speck of dust on a table in the house across the street.

4 0

3 years ago

The Surface Pressure at Leh, Ladakh is 800 mb. Now, assuming that Leh is at an altitude of 3500 m and every 100 m increase in he

Basile [38]

We have that the sea level pressure for Leh area is 1150mb mathematically given as

Ps= 1150 mb

<h3> Sea level pressure</h3>

Question Parameters:

Ladakh is 800 mb.

<u>assuming </u>that Leh is at an altitude of 3500 m and every 100 m

increase in height with respect to sea level corresponds to 10 mb pressure,

Generally, for 3500m the pressure change will be 350 mb.

Therefore, here for the sea level <em>pressure</em> we need to add,

Ps=800+350

Ps= 1150 mb

For more information on Pressure visit

brainly.com/question/25688500

8 0

2 years ago

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 32 km/h and the coef

Leviafan [203]

Answer: 29.83m

Explanation: coefficient of friction= v^2/rg

Coefficient of friction=0.27

V=32km/h

Convert km/h to m/s

32*1000/3600

32000/3600=8.89m/s

0.27=8.89^2/r*9.81

0.27*9.81*r=79.0321

R= 79.0321/2.6487

R=29.83m

5 0

3 years ago

Read 2 more answers

Un adolescente que va en monopatín rueda hacia abajo sobre un plano inclinado de 18.0 m de largo. El chico parte con una rapidez

qaws [65]

Answer: 31.62°

Explanation:

Tenemos como datos:

Distancia = 18.0m

Velocidad inicial = 2.0 m/s

Tiempo total = 3.3s

Sabemos que para un plano inclinado (ignorando el rozamiento) la aceleración se escribe como:

a(t) = g*sen(θ)

donde θ es el ángulo del plano inclinado, y g = 9.8m/s^2

Sabemos que para la velocidad tenemos que integrar la aceleración sobre el tiempo, entonces:

v(t) = g*sen(θ)*t + v0

Donde v0 es la velocidad inicial: v0 = 2.0m/s

v(t) = 9.8m/s^2*sen(θ)*t + 2.0m/s

Y para la posición, podemos integrar de vuelta sobre el tiempo:

p(t) = 0.5*9.8m/s^2*sen(θ)*t^2 + 2.0m/s*t + p0

Donde p0 es la posición inicial, podemos considerar que es cero para este problema.

p(t) = 4.9m/s^2*sen(θ)*t^2 + 2.0m/s*t

Y usando los datos iniciales, sabemos que en 3.3 segundos se recorren 18 metros, entonces:

p(3.3s) = 18m = 4.9m/s^2*sen(θ)*(3.3s)^2 + 2.0m/s*3.3s

18m = 51.744m*sen(θ) + 6.6m

sen(θ) = (18m - 6.6m)/ 51.744m

θ = cosec( (18m - 6.6m)/ 51.744m ) = 31.62°

4 0

3 years ago