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3 years ago

Displacement vectors of 4 km north, 2 km south, 5 km north, and 5 km south combine to a total displacement of

Physics

1 answer:

valentinak56 [21]3 years ago

6 0

south = -(north)

Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)

Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)

Displacement = (4 - 2 + 5 - 5) km north

<u>Displacement = 2 km north </u>

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the radio wave in this problem,

$\lambda = 0.3 m$

$c=300,000,000 m/s = 3\cdot 10^8 m/s$

Therefore, the frequency is:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz$

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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2 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

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3 years ago

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Firlakuza [10]

Answer:

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2 years ago