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3 years ago

What does the diagram represent?

Physics

1 answer:

Scrat [10]3 years ago

6 0

Answer:

Explanation:

the oxygen has a full electron shell, making it stable and neutral with no charges.

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Someone please help with this

SSSSS [86.1K]

Answer:

The new force is 2/3 of the original force

Explanation:

Coulomb's Law

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

$\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}$

Factoring out 2/3:

$\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}$

Substituting the original force:

$F'=\frac{2}{3}F$

The new force is 2/3 of the original force

7 0

3 years ago

Which conclusion about the atom can be drawn from Rutherford’s gold foil experiment?

Drupady [299]

D. Rutherford found that the atom consists of a small positively charged nucleus.

6 0

3 years ago

Read 2 more answers

What produces virtual and real images

german

Virtual images are produced when rays of a light (after reflection or refraction) appear to meet a a certain point. While real images are produced when rays of a light (after reflection or refraction) meet at some point.

3 0

4 years ago

In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a

SOVA2 [1]

Answer: $0.084 m/s^2$

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance $d_1=8.13 km$

time taken=25 min =1500 s

initial speed $v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s$

after 8.13 km mark steve started to accelerate

speed after 60 s

$v_2=v_1+at$

$v_2=5.6+a\times 60$

distance traveled in 60 sec

$d_2=v_1\times 60+\frac{a60^2}{2}$

$d_2=336+1800 a$

time taken in last part of journey

$t_3=1663.6-1560=103.6 s$

distance traveled in this time

$d_3=v_2\times t_3$

$d_3=\left ( 5.6+a\times 60\right )103.6$

and total distance $=d_1+d_2+d_3$

$10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6$

$a=0.084 m/s^2$

5 0

4 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Nezavi [6.7K]

Answer: 1.124 m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=2.80 m/s$ is the initial speed

$\theta=39\°$ is the angle at which the venom was shot

$t$ is the time since the venom is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.4 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin:

First we have to find $t$ from (2):

$0=0.4 m+2.8 m/s sin(39\°) t+\frac{-9.8m/s^{2}t^{2}}{2}$ (3)

Rearranging (3):

$-4.9 m/s^{2} t^{2} + 1.762 m/s t + 0.4 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: