WILL GIVE BRAINLIEST

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

$E=\dfrac{k}{d^2}$

We can write it as follows :

$\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2$

Put all the values,

$\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}$

So, the correct option is (a).

5 0

2 years ago

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest accelerating animals,

Andre45 [30]

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

$KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules$

Power

$P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W$

Converting to hp

$1\ W=\frac{1}{745.7}\ hp$

$\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp$

The power developed by the cheetah is 9241.6 W or 12.39318 hp

7 0

3 years ago

Compare animal and plant cell

vagabundo [1.1K]

Answer:

A plant cell contains a large, singular vacuole that is used for storage and maintaining the shape of the cell. In contrast, animal cells have many, smaller vacuoles. Plant cells have a cell wall, as well as a cell membrane. Animal cells simply have a cell membrane, but no cell wall.

hope this helps :)

8 0

2 years ago

Read 2 more answers

Kara is writing a science-fiction story. In the story, Earth no longer experiences day and night.

Ludmilka [50]

Answer:

C. She should get the Earth spinning on its axis again.

Explanation:

The Earth experiences day and night because of its spinning on its own axis.

If Kara's science-fiction story doesn't have the Earth spinning on its own axis, then the Earth will not experience day and night and hence Kara should incorporate this idea into his story.

6 0

3 years ago

Read 2 more answers

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago