1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
marta [7]
3 years ago
5

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 an

d d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth. The magnitude is N. The direction is °.
Physics
2 answers:
emmainna [20.7K]3 years ago
7 0
Here's the answer ladies and gents
Magnitude : 0.5n
Direction  : 0 degrees.
Marina CMI [18]3 years ago
5 0

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth.

The magnitude is  0.5 N.

The direction is °.0

ON EDGENUTITY

You might be interested in
List 3 examples in which friction helps us or makes things easier in our daily life. Explain the effect of friction for each.
Kisachek [45]

<span>Friction can help us play tennis, knock down bowling pins, and cook dinner.</span>

8 0
3 years ago
Read 2 more answers
A man, a distance d=3~\text{m}d=3 m from a target, throws a ball at an angle \theta= 70^\circθ=70 ​∘ ​​ above the horizontal. If
lbvjy [14]

Answer:

The ball doesn't strike the building because it strikes the ground at d=1.62 meters.

Explanation:

V= 5 m/s < 70º

Vx= 1.71 m/s

Vy= 4.69 m/s

h= Vy * t - g * t²/2

clearing t for the flying time of the ball:

t= 0.95 s

d= Vx * t

d= 1.62 m

4 0
3 years ago
Sam shared with his class that he lives on Oak Street. Sam is likely how old?
Lyrx [107]

D FOUR years old hope this helps


mark brainliest please

4 0
3 years ago
Read 2 more answers
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0
3 years ago
A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

6 0
3 years ago
Other questions:
  • (33%) Problem 3: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1100 kg and i
    6·1 answer
  • Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist
    6·1 answer
  • The shape of a sign gives you a clue about the information contained on the sign.
    13·1 answer
  • Conversion of the sun’s energy from fossil fuels into electricity does not involve:
    11·1 answer
  • why should it take significantly more energy to move a beam of alpha particles than a beam of beta minus particles
    9·1 answer
  • The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3 × 10−5 N · m2 /C when th
    11·1 answer
  • Help pleaseee is this correct?​
    14·1 answer
  • Calculate the wavelength λ and the frequency f of the photons that have an energy of E photon = 2.32 × 10 − 19 J. Ephoton=2.32×1
    14·1 answer
  • A fox locates its prey under the snow by slight sounds rodents make. The fox then leaps straight into the air, and then burrows
    13·1 answer
  • Is ironing clothes conduction, convection, or radiation?
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!