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2 years ago

Which is affected more by an earthquake, : deep sea or less deep sea???

Physics

2 answers:

lisabon 2012 [21]2 years ago

6 0

Answer:

deep sea will obviously be more affected because of the sea floor shaking

DaniilM [7]2 years ago

3 0

Answer:

deep see

because of the earth inside part highly pressure

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An ambulance is traveling north at 55.9 m/s, approaching a car that is also traveling north at 28.4 m/s. The ambulance driver he

wlad13 [49]

Answer:

915 Hz

Explanation:

The observed frequency from a sound source is given as

f₀ = f [(v + v₀)/(v+vₛ)]

where

f₀ = observed frequency of the sound by the observer = ?

f = actual frequency of the sound wave = 983 Hz

v = actual velocity of the sound waves = 343 m/s

vₛ = velocity of the source of the sound waves = 55.9 m/s

v₀ = velocity of the observer = 28.4 m/s

f₀ = 983 [(343+28.4)/(343+55.9)]

f₀ = 915.2 Hz = 915 Hz

6 0

3 years ago

What is principal focus and focus length with examples

Shkiper50 [21]

Explanation:

Principle focus is the point on the axis of a convex lens, where the parallel rays of light from one side of the lens. meet on other side after refraction. Distance between optical centre to principle focus point is the focal length.

7 0

2 years ago

Consider a standing wave on a string. What is the distance between two adjacent nodes in terms of the wavelength λ of the standi

amid [387]

The distance between the two adjacent nodes = λ/2.

<h3>What is Wavelength?</h3>

A periodic wave's wavelength is its spatial period, or the length over which its form repeats. It is a property of both travelling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda (λ) is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

The distance between the two adjacent nodes = λ/2.

for the standing wave ,the distance between any two adjacent nodes or antinodes is 1/2 λ.

to learn more about the wavelength go to - brainly.com/question/6297363

#SPJ4

3 0

2 years ago

The relationship between linear velocity and angular velocity

bonufazy [111]

In linear motion , when a body moves with uniform velocity , in time t , its linear displacement will be ;

S = r∅ S = vt

r∅ = vt

r.∅ / t = v

v = rw

where ∅ = 90° is the angle between between radius vector r and angular velocity w (omega )

In case ∅ ≠ 90° , we can write v = r w sin∅

It gives us v = w× r

7 0

2 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

8 0

3 years ago