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3 years ago

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Betty is sitting on of her surfboard out in the ocean. She is waiting for the perfect wave to come along so she can ride in it t

o shore. As she waits, she notices that the waves roll by in pattems, or sets. As the top of each wave passes by Betty, it pushes her up. Which part of the wave does this?
a. compression
b. Crest
c. rarefaction
d. trough

1 answer:

Ahat [919]3 years ago

7 0

Answer:

the type of wave is a crest

Explanation:

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates

Alina [70]

Answer:

The capacitance of the deflecting plates is $C=1.59 pF$ .

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

$C=\frac{\varepsilon _{0}A}{d}$

Here, C is the capacitance, A is the area, d is the distance and $\varepsilon _{0}$ is the absolute permittivity.

Convert the side of the square from cm to m.

s= 3.0 cm

s= 0.030 m

Calculate the area of the square.

$A= s_^{2}$

Put s= 0.030 m.

$A=(0.030)_^{2}$

$A=9\times10^{-4}m^{-2}$

Convert distance from mm to m.

d= 5.0 mm

$d=5\times10^{-3}m$

Calculate the capacitance of the deflecting plates.

$C=\frac{\varepsilon _{0}A}{d}$

Put $d=5\times10^{-3}m$ , $A=9\times10^{-4}m^{-2}$ and $\varepsilon _{0}=8.85\times 10^{-12}Fm^{-1}$ .

$C=\frac{(8.85\times 10^{-12})(9\times10^{-4})}{5\times10^{-3}}$

$C=1.59\times 10^{-12}F$

$C=1.59 pF$

Therefore, the capacitance of the deflecting plates is $C=1.59 pF$ .

4 0

3 years ago

In addition to plate tectonics, what is the other main driving force of the rock cycle?

GREYUIT [131]

Answer: Option B

Explanation:

Rocks contains pores and spaces in them in which water gets trapped, and this causes the rock to break as water increases its volume due to freezing. This rocks are then eroded and are carried away into the streams and channels. The rivers and streams can carry particles of various shapes and size, which are further transported and deposited at a different place.

This particles or sediments then gets compacted and lithified in due course of time, and forms sedimentary rocks.

Water is also needed to undergo metasomatism process, that creates a metamorphic rock.

Hence, water flow is also an important driving force in rock cycle, along with tectonic activities.

Thus, the correct answer is option (B).

4 0

3 years ago

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An object at free fall near the Earth’s surface experiences constant velocity.

mariarad [96]

False it accelerates.

7 0

3 years ago

There is a current I flowing in a clockwise direction in a square loop of wire that is in the plane of the paper. If the magneti

nevsk [136]

Answer:

Explanation:

The magnetic moment of the coil M = current x area of loop

= I x L x L

M = L² I .

Magnetic field = B

Torque = M x B

= MB sinθ where θ is angle between M and B . Direction of M is perpendicular to face of loop

Torque = MB sin90

= MB

= L² I B .

3 0

3 years ago