。and 。 are needed to describe a force

nikdorinn [45]

Answer:

As the football travels up, gravity slows it down until it reaches a point where it stops briefly at its highest peak; then the football comes down, and gravity accelerates it until it hits the ground hard.

Have a great day!

8 0

3 years ago

A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn

umka2103 [35]

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

$E=\dfrac{kq}{r^2}$

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

5 0

4 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

How do you know the speed of an electromagnetic wave in vacuum?

larisa [96]

Answer:

when huburxvkruxv hfgy

8 0

3 years ago

Read 2 more answers

A book that weighs 16N is on top of a shelf that is 3m high. What is the work required to

DENIUS [597]

48J

Explanation:

Given parameters:

Weight of the book = 16N

Height of the shelf = 3m

Unknown:

Work done to raise the book = ?

Solution:

Work done is defined as the product of force and distance. The force here is the weight of the book which signifies that it was carried in the vicinity of gravitational force.

Work done = Force x distance = weight x distance

Weight is a type of force;

Input the parameters;

Work done = 16 x 3 = 48J

The unit of workdone is in Joules

learn more:

Work done brainly.com/question/9100769

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8 0

3 years ago