6 An object can never have

Only question 7. I keep getting it wrong and i’m not sure

Akimi4 [234]

Answer:

I think you just have to input the horizontal velocity component.

Explanation:

Assuming no air resistance

The horizontal velocity is constant at

vx = 17.5cos71 = 5.6974... = 5.70 i m/s

The vertical velocity varies with gravity

vy = 17.5sin71 + -9.81(22) = -199.273... = -199 j m/s

v = 5.70i - 199j m/s

arguably one should round to only two significant digits

3 0

3 years ago

A force f = bx 3 acts in the x direction, where the value of b is 3.7 n/m3. how much work is done by this force in moving an obj

Elodia [21]

The answer would be 21.6 but rounded up it would be 22J.

8 0

3 years ago

Two speakers placed 0.94 m apart produce pure tones in sync with each other at a frequency of 1630 Hz. A microphone can be moved

Alina [70]

Answer:

a. approximately $1.1\; \rm m$ (first minimum.)

b. approximately $2.2\; \rm m$ (first maximum.)

c. approximately $3.4\; \rm m$ (second minimum.)

d. approximately $4.7\; \rm m$ (second maximum.)

Explanation:

Let $d$ represent the separation between the two speakers (the two "slits" based on the assumptions.)

Let $\theta$ represent the angle between:

the line joining the microphone and the center of the two speakers, and
the line that goes through the center of the two speakers that is also normal to the line joining the two speakers.

The distance between the microphone and point $P_0$ would thus be $9.4\, \tan(\theta)$ meters.

Based on the assumptions and the equation from Young's double-slit experiment:

$\displaystyle \sin(\theta) = \frac{\text{path difference}}{d}$ .

Hence:

$\displaystyle \theta = \arcsin \left(\frac{\text{path difference}}{d}\right)$ .

The "path difference" in these two equations refers to the difference between the distances between the microphone and each of the two speakers. Let $\lambda$ denote the wavelength of this wave.

$\displaystyle \begin{array}{c|c} & \text{Path difference} \\ \cline{1-2}\text{First Minimum} & \lambda / 2 \\ \cline{1-2} \text{First Maximum} & \lambda \\\cline{1-2} \text{Second Minimum} & 3\,\lambda / 2 \\ \cline{1-2} \text{Second Maximum} & 2\, \lambda\end{array}$ .

Calculate the wavelength of this wave based on its frequency and its velocity:

$\displaystyle \lambda = \frac{v}{f} \approx 0.211\; \rm m$ .

Calculate $\theta$ for each of these path differences:

$\displaystyle \begin{array}{c|c|c} & \text{Path difference} & \text{approximate of $\theta$} \\ \cline{1-3}\text{First Minimum} & \lambda / 2 & 0.112 \\ \cline{1-3} \text{First Maximum} & \lambda & 0.226\\\cline{1-3} \text{Second Minimum} & 3\,\lambda / 2 & 0.343\\ \cline{1-3} \text{Second Maximum} & 2\, \lambda & 0.466\end{array}$ .

In each of these case, the distance between the microphone and $P_0$ would be $9.4\, \tan(\theta)$ . Therefore:

At the first minimum, the distance from $P_0$ is approximately $1.1\; \rm m$ .
At the first maximum, the distance from $P_0$ is approximately $2.2\; \rm m$ .
At the second minimum, the distance from $P_0$ is approximately $3.4\; \rm m$ .
At the second maximum, the distance from $P_0$ is approximately $4.7\;\rm m$ .

6 0

4 years ago

A 0.155 kg arrow is shot upward

Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0

3 years ago

Select three different examples of accelerated motion. a body traveling in a straight line and increasing in speed a body travel

IrinaVladis [17]

This is the same question that I just answered.

Have present the definition of acceleration:

a = Δv / Δt, this is change in velocity per unit of time.

a and v are in bold to mean that they are vectors.

1) a body traveling in a straight line and increasing in speed: CORRECT:

Acceleration is the change in velocity, either magnitude or direction or both. So, a body increasing in speed is accelerated.

2) a body traveling in a straight line and decreasing in speed: CORRECT

A decrease in speed is a change in velocity, so it means acceleration.

3) a body traveling in a straight line at constant speed: FALSE.

That body is not changing either direction or speed so its motion is not accelerated but uniform.

4) a body standing still : FALSE.

That body is not changind either direction or speed.

5) a body traveling at a constant speed and changing direction: CORRECT.

The change in direction means that the body is accelerated. The acceleration due to change in direction is named centripetal acceleration.

6 0

4 years ago