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larisa86 [58]
3 years ago
5

Minute after minute, hour after hour, day after day, ocean waves continue to splash onto the shore. Explain why the beach is not

completely submerged and why the middle of the ocean has not yet been depleted of its water supply. A) Ocean waves can only bring water to the shore but cannot take it back. The momentum of the water particles switches direction when the wave hits the shore and this momentum takes all the water back into the ocean. B) Ocean waves bring the water and also take it all the way back into the ocean. As such, water does not pile up on the beach. C) Ocean waves can only bring energy to the shore; the particles of the medium (water) simply oscillate about their fixed position. D) None of the choices are correct.
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Answer: C. Ocean waves can only bring energy to the shore; the particles of the medium (water) simply oscillate about their fixed position.

Explanation:

The reason why the beach is not completely submerged and the reason why the middle of the ocean has not been depleted of its water supply is due to the fact that water isn't transported by ocean waves.

It should be noted that even a single drop of water cannot be brought by the ocean wave to the shore from the middle of the ocean.

The only thing that the ocean waves can bring to the shore is energy. Hemce, the water particles oscillate in their fixed position which is vital in making sure that the beach isn't piled up with water.

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A hollow conductor is positively charged. Asmall uncharged metal ball is lowered by a silk thread through asmall opening in the
dimulka [17.4K]

Answer:

Explanation:

According to the property of a conductor, the entire charge will reside on the outer surface of the conductor, there is no charge on the inner side of the conductor. As the uncharged metal ball touches the inner surface of the conductor, it does not attain any charge as the inner side of the conductor has no charge.

So option (c) is correct.

8 0
3 years ago
A diffraction grating contains 15,000 lines/inch. We pass a laser beam through the grating. The wavelength of the laser is 633 n
MatroZZZ [7]

Answer:

Recall the Diffraction grating formula for constructive interference of a light

y = nDλ/w                                      Eqn 1

Where;

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10-9 m  

and n = 1  

y₁ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1)                =         0.994m

Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?

w = width of slit = 1/15000in =6.67x10⁻⁵in =   6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m

D = distance to screen  

λ = wavelength of light  

n = order number  = 1

Given  

y1 = ? from 1st order max to the central  

D = 2.66 m  

λ = 633 x 10⁻⁹ m  

and n =  2

y₂ = 0.994m  

Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m

8 0
3 years ago
A combined circuit has two resistors in series (27.5 ohms and 33.0 ohms) and another in parallel (22.0 ohms). If the power sourc
Dafna1 [17]
R 1,2 = 27.5 + 33.0 = 60.5 Ohms
1/ R 1,2,3 = 1/ 60.5 + 1 / 22 = 82.5 / 1331
R 1, 2, 3 = 1331 / 82.5 = 16.13 Ohms
I = U / R
I = 9 V / 16.13 Ohms = 0.557 A ≈ 0.56 A
Answer: C ) 0.56 Amps 
4 0
3 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
Vector B has x y, and z components of 8.8,
Ostrovityanka [42]

Answer: 14.122 units

Explanation:

We have the following vector:

\vec{V}=(x,y,z)=(8.8, 7.4, 8.2)

If we want to find its magnitude we have to use the following formula:

V=\sqrt{x^{2}+y^{2}+z^{2}}

V=\sqrt{(8.8units)^{2}+(7.4units)^{2}+(8.2 units)^{2}}

V=14.122 units  This is the magnitude of vector \vec{V}

3 0
3 years ago
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