Ask question

Ask question

All categories

3 years ago

7

a uniform electric field E=15 N/C points downwards. A particle with charge q=-0.15 C is placed in the electric field. What is th

e magnitude and direction of the force on the particle?

1 answer:

Alex17521 [72]3 years ago

3 0

Answer:

2.25 N upwards

Explanation:

Answer:

800 N/C to the right

Explanation:

The equation that relates force on a charge, electric field and charge is

$F=qE$

where

F is the force acting on the charge

q is the charge

E is the electric field

In this problem, we have

$E=15 N/C$ is the electric field

$q=-0.15 C$ is the charge

Substituting into the formula, we find the force

$F=(-0.15 C)(15 N/C)=-2.25 N$

Concerning the direction:

- The electric field and the force have same directions if the charge is positive

- The electric field and the force have opposite directions if the charge is negative

Here the charge is negative, so the electric field has opposite direction to the force: therefore, it must be upwards.

You might be interested in

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago

Select the correct arrow.<br> Which arrow correctly shows the flow of heat?<br> Reset<br> Next

julsineya [31]

Answer:

where is the image?

Explanation:

6 0

2 years ago

Read 2 more answers

Match the terms to the correct descriptions. Question 1 options:

faust18 [17]

1) Refraction
2)Reflection
3)Concave
4)Convex

I took the test and got this right so you can believe me :)
Hope this helps

4 0

3 years ago

Read 2 more answers

A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

astraxan [27]

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$

7 0

3 years ago

A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v

Pavel [41]

Answer:

65

Explanation:

The resonant frequencies for a fixed string is given by the formula nv/(2L).

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e fn+1 – fn = 390 -325

= 65

3 0

3 years ago