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2 months ago

15

Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an

d why

1 answer:

Keith_Richards [3.2K]2 months ago

7 0

Both students are correct because they highlight different aspects of waves. Raphael describes the wave in terms of its wavelength, while Lucinda focuses on its frequency, showing that both measurements are valid.

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A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad

Keith_Richards [3271]

Answer:

9.21954 m/s

54 m/s²

The angle is zero

Explanation:

r = Length of the arm = 1.5 m

$\omega$ = Angular velocity = 6 rad/s

The speed's horizontal component can be calculated using

$v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s$

The vertical speed is determined by

$v_v=2\ m/s$

The combination of these two components yields the hammer’s speed in relation to the ground

$v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s$

The hammer’s speed relative to the ground measures 9.21954 m/s

Acceleration in the vertical direction is zero

The total acceleration is expressed as

$a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2$

Total acceleration is 54 m/s²

Since acceleration points towards the center, the angle remains zero.

3 0

2 months ago

Subsurface ocean currents continually circulate from the warm waters near the equator to the colder waters in other parts of the

ValentinkaMS [3465]

<span>The primary factor behind these currents is the </span>variability in the density of ocean waters.

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1 month ago

A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

Yuliya22 [3333]

To tackle this problem, it's essential to employ concepts associated with force as per Hooke's law, alongside the forces described by Newton's second law and the concept of potential elastic energy. Since the forces are in equilibrium, the spring force matches the gravitational force. To find the spring constant k, we recognize the compression is 40cm at launch, hence applying the potential elastic energy formula results in determining the energy stored in the spring as 63.72 Joules.

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1 month ago

A 3030 cmcm wrench is used to loosen a bolt with a force applied 0.30.3 mm from the bolt. It takes 6060 NN to loosen the bolt wh

Softa [3030]

Complete Question

A 30 cm cm wrench is employed to loosen a bolt, with force applied 0.30m from the bolt. It requires 60 N to loosen the bolt when force is applied perpendicular to the wrench. How much force would be necessary if the force were applied at a 30-degree angle from perpendicular?

Response:

The strength needed is $F_{\theta } = 69.28 \ N$

Clarification:

According to the problem, we know that

The length of the wrench is $L = 30 cm = \frac{30}{100} = 0.3 \ m$

The distance from the bolt is $d = 0.30 m$

The force necessary to loosen the bolt is $F = 60 N$

The angle of application is $\theta = 30 ^o$

Generally, the torque required for loosening the bolt is defined as

$\tau = F * d$

$\tau = 60 * 0.3$

$\tau = 18 Nm$

Now for the bolt to loosen at $\theta$ the torque at 90° must equal that at $\theta$

Thus, the torque at $\theta$ is represented mathematically as

$\tau = F_{\theta }d cos \theta$

substituting values gives us

$18 = F_{\theta } * 0.3 cos (30)$

$F_{\theta } = \frac{18}{0.3 cos (30)}$

$F_{\theta } = 69.28 \ N$

7 0

1 month ago

A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

ValentinkaMS [3465]

The ball was released from a height of 20 meters

Explanation:

The scenario is as follows:

1. A ball drops from the edge of a cliff.

2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.

This implies K.E = P.E.

3. The ball impacts the ground at a speed of 20 m/s.

4. The gravitational field strength noted is 10 N/kg.

<pOur goal is to ascertain the height from which the ball was dropped.

<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final velocity, $v_{0}$ is the initial velocity, and m is the mass.

<p→ v = 20 m/s and $v_{0}$ = 0 m/s.<p→ K.E = $\frac{1}{2}m(20^{2}-0^{2})$

→ K.E = $\frac{1}{2}m(400)$

→ K.E = 200 m joules when the ball strikes the ground.

<p→ P.E = mg h

where g is the gravitational field strength, m is mass, and h signifies height.

<p→ g = 10 N/kg.<p→ P.E = m(10)(h)

→ P.E = 10m h joules.

<p→ P.E = K.E.

→ 10m h = 200 m.

Dividing through by 10m yields:

→ h = 20 meters.

The ball was released from a height of 20 meters.

Learn more

To understand more about gravitational potential energy, visit

8 0

2 months ago