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2 years ago

7

An oil droplet is levitated and held at rest in a region where the electric field is 1.4x104 N/C directed vertically downwards.

If the droplet has 3 excess electrons attached, what is its mass

1 answer:

Ratling [72]2 years ago

8 0

W = m g weight of drop

W = q E

m = q E / g = = 3 e E / g

m = 3 * 1.6E-19 * 1.4E4 / 9.8 = 6.96E-16 kg

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Solenoid 2 has twice the radius and six times the number of turns per unit length as solenoid 1. The ratio of the magnetic field

Xelga [282]

Answer:

6

Explanation:

The magnetic field inside a solenoid is given by the following formula:

$B = \mu_{0}nI$

where,

B = Magnetic Field Inside Solenoid

μ₀ = permittivity of free space

n = No. of turns per unit length

I = Current Passing through Solenoid

For Solenoid 1:

$B_{1} = \mu_{0}n_{1}I ------------------- equation 1$

For Solenoid 2:

n₂ = 6n₁

Therefore,

$B_{1} = \mu_{0}n_{2}I\\B_{1} = 6\mu_{0}n_{1}I ----------------- equation 2$

Diving equation 1 and equation 2:

$\frac{B_{2}}{B_{1}} = \frac{6\mu_{0}nI}{\mu_{0}nI}\\\\\frac{B_{2}}{B_{1}} = 6$

Hence, the correct option is:

<u>6</u>

8 0

2 years ago

A cheetah can go from the state of rest to running at 20m/s in just two seconds. What is the Cheetahs average acceleration

babymother [125]

acceleration = change in velocity/change in time

so...

a = 20 m/s / 2 seconds

a = 10

hope that helps :)

P.S. found this from Brainly User, sometimes all you have to do is search to find the answer.

7 0

2 years ago

2. Draw a ray diagram that shows the image

Lunna [17]

you have to draw diagram

6 0

2 years ago

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup

Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

$x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km$

Then the ship traveled $384,000-9,076.9104*2 = 361,846.1792km$ at constant speed, which means that it traveled:

$f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95$

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

$t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min$

That added to the other interval times:

$t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min$

5 0

2 years ago

Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed

ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance = 29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0

2 years ago