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zysi [14]
3 years ago
7

During a visit to the beach, you get in a small rubber raft and paddle out beyond the surf zone. Tiring, you stop and take a res

t. Describe the movement of your raft during your rest. How does this movement differ, if at all, from what you would have experienced if you had stopped paddling while in the surf zone?
Physics
1 answer:
Monica [59]3 years ago
3 0

Answer:

The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach

Explanation:

The movement in the two parts is very different, when the surf zone has passed it is in a deeper part of the water where the seabed does not rise much, therefore due to the movement of the waves there is an upward oscillatory movement and descending, in this movement there is no horizontal displacement.

When it is within the southern zone, there is a rapid rise of the sea floor, which generates a horizontal movement, having a traveling wave, therefore your movement is more complicated, you can have some oscillating movement on the axis and, but in addition to this you have a horizontal movement that reaches you towards the beach, forming a Traveling wave.

The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach

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Why would the bulb not light?
Norma-Jean [14]

Im not sure this is correct but, i think thats because the wire connected at the negative side. The current go from positive to negative and thats why the bulb not light, also the bulb location is not at the center of the battery which explain why there's no current at the upper side.

Hope you understand. Sorry if my english sucks YEET

6 0
3 years ago
Read 2 more answers
All of the following steps are included in a self-modification program except __________. A. specifying target behavior B. desig
Y_Kistochka [10]

Answer:

C)focusing on one goal at a time

Explanation:

Self-modification programs could be regarded as a program that help individual in managing unwanted as well as dysfunctional behavioral responses whenever they are going through a problem and try to deal with it. For instance the dysfunctional behavioral response for someone with a panic attack is avoidance. Some of the the steps that are involved in in a self-modification program are;

✓ specifying target behavior ✓designing a program

✓ gathering data about target behavior

8 0
3 years ago
Read 2 more answers
A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red
Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }       ......(1)

Here \lambda_{s} is source wavelength, \lambda_{o} is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, \lambda_{s} = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, \lambda_{0} = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }

\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84

{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056

1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}

\frac{v}{c}=\frac{0.2944}{8.056}

v = 0.036c=0.036\times3\times10^{8}

v = 1.08 x 10⁷ m/s  

8 0
3 years ago
A stone is thrown vertically upward with an initial speed of 24.7 m/s. Neglect air resistance. The speed of the stone when it is
fredd [130]

Answer:

The speed of the stone when it is 4.66 m higher is 236.057 m/s.

Explanation:

Given the initial velocity and vertical distance, we can use the fourth kinematic equation (v^{2} =v_{o}^{2}+2ay) to find v final, or the v to the left of the equal sign. We know v_{o} (initial velocity) is 24.7 m/s, y (change in vertical distance) is 4.66 m, and a is another way to write g (acceleration due to gravity), or 9.8 m/s^{2}.

From here you could plug in the values and solve for v final, but to make the solving process simpler, we can simplify the given equation, <em>then </em>plug in the known values.

To isolate v final, we can take the square root of v^{2} and do the same to the right side of the equation. Therefore, we can find v final with: v_{o} \sqrt{2ay}, where v initial is outside of the square root because it squared...

If we plug in the known values to the simplified equation, we get: v=24.7m/s*\sqrt{2(9.8m/s)(4.66 m)}

The final answer is 236.057 m/s.

4 0
2 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
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