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3241004551 [841]

3 years ago

10

Radar can be used to detect the location and movement of precipitation. Please select the best answer from the choices provided

T F

1 answer:

sveta [45]3 years ago

3 0

It is true that radar can be used to detect the location and movement of precipitation.

You might be interested in

A material you are testing conducts electricity but canot be pulled into wires

agasfer [191]

A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid. Hope this helps!

4 0

3 years ago

Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r

cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

E_net = E_2 + E_4

E_2 = E_4

E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

5 0

3 years ago

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago

Wilma can mow a lawn in 80 minutes. Rocky can mow the same lawn in 120 minutes. Construct an equation that would allow you to de

xeze [42]

Answer:

$t = 96 minutes$

Explanation:

Time to mow 1 lawn by Wilma is 80 minutes

so work done in 1 minute by Wilma is given as

$W_1 = \frac{1}{80}$

Similarly Rocky mow same lawn in 120 minute

so work done in 1 minute by Rocky is given as

$W_2 = \frac{1}{120}$

now we know that they both worked by "t" time

so total work performed by them

$W = \frac{t}{80} + \frac{t}{120}$

they both mow 2 lawns then it is given as

$2 = (\frac{1}{80} + \frac{1}{120})t$

$t = 96 minutes$

5 0

3 years ago

Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillat

ozzi

Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

<u>wavelength = 24 m</u>

Period:

The period is given as:

$Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\$

<u>Period = 10 s</u>

<u></u>

Frequency:

The frequency is given as:

$f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\$

<u>f = 0.1 Hz</u>

<u></u>

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

<u>Amplitude = 4 m</u>

5 0

3 years ago