All categories

3 years ago

Compare how a positively charged object and a negatively charged object interact with a neutral object

2 answers:

3 0

A neutral is simply one which contains and equal number of positive and negative charges. So when a charged object interacts with the neutral object, it attracts the opposite charge to the side of the interaction and repels the same charge to the opposing side of the interaction. <span>+ object<span>- object</span>
</span>

Natali5045456 [20]3 years ago

3 0

Well, as far as I know, the charged objects don't interact with "neutral" objects. Unless the charged objects induce charges in the otherwise neutral objects ...

You might be interested in

9.80 1. The density of mercury is 13600 kg/m³. What is this value in g/cm³? 2. Find the mass of water which will fit in a large

My name is Ann [436]

Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. $density=\frac{mass}{volume}$

$1000=\frac{mass}{2 * 1 * 0.2}$

$1000*0.4=mass$

$400kg = mass$

3. $density=\frac{mass}{volume}$

$0.6=\frac{120}{volume}$

$volume=\frac{120}{0.6}$

$volume= 200cm$

4 0

2 years ago

How often does a 16-year-old need a medical evaluation from a doctor?

a_sh-v [17]

Answer:

Explanation: A

5 0

3 years ago

Read 2 more answers

A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A

Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

$IMA = \frac{d_i}{d_o}$

where

$d_i$ is the input distance

$d_o$ is the output distance

For the pulley system in this problem, $d_i = 3.90 m$ and $d_o = 0.975 m$ , so the IMA is

$IMA=\frac{3.90 m}{0.975 m}=4$

B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

$MA=\frac{F_o}{F_i}$

where $F_o$ is the output force and $F_i$ is the input force. For the pulley system in this problem, $F_i = 375 N$ and $F_o = 1345 N$ , so the MA is

$MA=\frac{1345 N}{375 N}=3.59$

C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

$\eta = \frac{MA}{AMA} \cdot 100$

Therefore, in this case,

$\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%$

3 0

3 years ago

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. how much time does it ta

Yuri [45]

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).

When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.

Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.

The equation of motion is stated as,

v = u + at

v² = u² + 2as

A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.

Then the time taken by the car will be,

u = 12 m/s

v = 18 m/s

s = 60 m

Put these in the equation v² = u² + 2as.

18² = 12² + 2 x a x 60

a = 1.5

Then the time will be

18 = 12 + 1.5t

1.5t = 6

t = 4 seconds

Hence, the time taken is 4 s.

The complete question is:

While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?

1.00 s

2.50 s

4.00 s

4.50 s

To know more about acceleration refer to: brainly.com/question/12550364

#SPJ4

8 0

1 year ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

3 years ago