A wave has a wavelength of 3 m and a frequency of 340 Hz. How far will the wave go in 20 s?

Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

2 years ago

In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

kondaur [170]

The moment of inertia is $24.8 kg m^2$

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

$I_r = \frac{1}{12}ML^2$

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

$I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2$

The moment of inertia of one ball is given by

$I_b = mr^2$

where

m = 50 kg is the mass of the ball

$r=\frac{L}{2}=\frac{0.96}{2}=0.48 m$ is the distance of each ball from the axis of rotation

So we have

$I_b = (50)(0.48)^2=11.5 kg m^2$

Therefore, the total moment of inertia of the system is

$I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2$

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

6 0

3 years ago

When the heart beats faster, blood vessels need to _____ more.<br> open <br> close move

Yuki888 [10]

I'm pretty sure it is open more

6 0

3 years ago

Read 2 more answers

Heat is transferred from the sun to the earth via electromagnetic waves (see Chapter 24). Because of this transfer, the entropy

Sever21 [200]

Answer:

1. Decreases,

2. Increases,

3. Increases

Explanation:

The heat which is a product of sun's energy, is transferred from the sun to the earth through radiation, conduction or convention. This heat passes through the earth atmosphere, then warms it , before becoming heat energy.

Therefore, Heat is transferred from the sun to the earth via electromagnetic waves . Because of this transfer, the entropy of the sun DECREASES, the entropy of the earth INCREASES and the entropy of the sun-earth system INCREASES.

8 0

3 years ago

What electrons are in “motion”, what do you have?

Sergeeva-Olga [200]

Answer:

When a positive charged object is placed near a conductor electrons are attracted the the object. ... When electric voltage is applied, an electric field within the metal triggers the movement of the electrons, making them shift from one end to another end of the conductor. Electrons will move toward the positive side. As you know, electrons are always moving. They spin very quickly around the nucleus of an atom. As the electrons zip around, they can move in any direction, as long as they stay in their shell.

7 0

3 years ago