What is the medium for this wave?

A 63.3 kg wood board is resting on very smooth ice in the middle of a frozen lake. A 35.7 kg boy stands at one end of the board.

Vlada [557]

The velocity of the board relative to the ice is zero, since both are at rest.

<h3>What is relative velocity?</h3>

Relative velocity is the velocity of an object in relation to another reference object or point.

When two objects are travelling or moving with the same velocity in the same direction, the relative velocity one relative to the other is zero.

Also, when two objects are at rest, the relative velocity one relative to the other is zero.

Therefore, the velocity of the board relative to the ice is zero, since both are at rest.

Learn more about relative velocity at: brainly.com/question/24337516

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8 0

2 years ago

Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take

KiRa [710]

Answer:

$\frac{V_{e}}{V_{h}}=0.428*10^{2}$

Explanation:

From conservation of energy states that

$K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}$

5 0

3 years ago

Which tool is used to track which organisms are carriers of a specific trait through several generations?

maw [93]

Answer:

answer is a pedigree chart :)

Explanation:

5 0

3 years ago

Read 2 more answers

A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by

Marizza181 [45]

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

$f_o$ = Focal length of objective = 0.9 cm

$f_e$ = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

$m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52$

The angular magnification of the compound microscope is -252.52

6 0

3 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

2.

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$

4 0

2 years ago