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juin [17]
3 years ago
12

The diagram shows a wave traveling through a medium.

Physics
2 answers:
m_a_m_a [10]3 years ago
6 0
The answer is b. The crest.

Hope this helps!
kakasveta [241]3 years ago
5 0

Answer:

Crest

For your question "The diagram shows a wave traveling through a medium.

An illustration of a transverse wave with the high point of the wave labeled B, the low point of the wave labeled C, the height of the wave was labeled A and the distance between two adjacent low points labeled D.

Identify point B.

amplitude

crest

trough

wavelength"

the answer is crest

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A series circuit contains a 9-volt battery, a 3-ohm resistor and a 2-ohm resistor. What is the voltage drop across the 2-ohm res
BARSIC [14]
Since everything in the circuit is in series .. .

-- The total resistance is  (3 + 2) = 5 ohms.

-- The voltage across the 3-ohm resistor is 3/5 of the total voltage.

-- The voltage across the 2-ohm resistor is 2/5 of the total voltage.

                  (2/5) of (9 volts)  =  18/5  =  3.6 volts .

7 0
3 years ago
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The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.
Lesechka [4]

The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it. Because the contact between the bolt and the   pliers working surface is less.

<h3>What is mechanical advantage ?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of the given mechanical machine.

The efficiency to open the stubborn bolt depends upon the contact between the working surface of the pliers and the bolt.

The contact between the bolt and the top pair of pliers working surface is less. Its mechanical advantage is less.

Hence, the top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.

To learn more about the mechanical advantage, refer to the link;

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3 0
2 years ago
Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football
faltersainse [42]

Answer:

a=159.32\ m/s^2

Explanation:

It is given that,

Angular speed of the football spiral, \omega=6.9\ rev/s=43.35\ rad/s

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

v=r\omega

v=0.085\times 43.35

v = 3.68 m/s

The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(3.68)^2}{0.085}

a=159.32\ m/s^2

So, the centripetal acceleration of the laces on the football is 159.32\ m/s^2. Hence, this is the required solution.

6 0
3 years ago
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
3 years ago
Humpback whales are known to produce a collection of elaborate and repeating sounds with frequencies starting at 20 Hz. The soun
jasenka [17]

Answer:

70 m.

Explanation:

Given,

Frequency, f = 20 HZ

speed of sound, v = 1400 m/s

wavelength of the waves = ?

we know,

v = f λ

\lambda= \dfrac{v}{f}

\lambda= \dfrac{1400}{20}

\lambda=70\ m

Hence, the wavelength of the wave is equal to 70 m.

8 0
3 years ago
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