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NARA [144]
3 years ago
5

Fill in the blanks. The electrostatic force between two objects is proportional to the ____________________ of the distance ____

________________.
Physics
2 answers:
Shkiper50 [21]3 years ago
6 0
Write an equation to calculate the force between two objects if the product of their charges is 10.0 × 10-4 C. (Note: Use the variable R for the distance between the charges.)

F = 900 ÷_________
Luba_88 [7]3 years ago
6 0

Answer:

The electrostatic force between two objects is proportional to the inverse of the square of the distance between the centres of the two objects

Explanation:

According to Coulomb's law of electrostatic attraction, the electrostatic force F between two charges or objects is directly proportional to the product of the absolute values of the charges or (charges on the objects) and inversely proportional to the square of their distance of separation.

Given that q_1 and q_2 are the charges or the charges on the objects, we can break down Coulomb's statement mathematically as follows;

F\alpha q_1q_2.............(1)  F is proportional to the product of the charges.

F\alpha \frac{1}{r^2}.......................(2) F is proportional to the inverse of the square of their distance r of separation.

This is an inverse-square relationship.

You might be interested in
One strategy in a snowball fight is to throw
faltersainse [42]

Answers:

a) \theta_{2}=23\°

b) t=1.199 s

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

x=V_{o}cos\theta_{1} t_{1}   (1)

Where:

V_{o}=11.1 m/s is the initial speed  of snowball 1 (and snowball 2, as well)

\theta_{1}=67\° is the angle for snowball 1

t_{1} is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}   (2)

Where:

y_{o}=0  is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

y=0  is the final height of the  snowball 1

g=-9.8m/s^{2}  is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

x=V_{o}cos\theta_{2} t_{2}   (3)

Where:

\theta_{2} is the angle for snowball 2

t_{2} is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}   (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate t_{1} from (2):

0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}   (5)

t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}   (6)

Substituting (6) in (1):

x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})   (7)

Rewritting (7) and knowing sin(2\theta)=sen\theta cos\theta:

x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})   (8)

x=-\frac{(11.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(67\°))   (9)

x=9.043 m   (10)  This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate t_{2} from (4) and substitute it on (3):

t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}   (11)

x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})   (12)

Rewritting (12):

x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})   (13)

Finding \theta_{2}:

2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})   (14)

2\theta_{2}=45.99\°  

\theta_{2}=22.99\° \approx 23\°  (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle (\theta_{2} < \theta_{1}).

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of t_{1} and t_{2} from (6) and (11), respectively:

t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}  

t_{1}=-\frac{2(11.1 m/s)sin(67\°)}{-9.8m/s^{2}}   (16)

t_{1}=2.085 s   (17)

t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}  

t_{2}=-\frac{2(11.1 m/s)sin(23\°)}{-9.8m/s^{2}}   (18)

t_{2}=0.885 s   (19)

Since snowball 1 was thrown before snowball 2, we have:

t_{1}-t=t_{2}   (20)

Finding the time difference t between both:

t=t_{1}-t_{2}   (21)

t=2.085 s - 0.885 s  

Finally:

t=1.199 s  

7 0
3 years ago
PART ONE
Helga [31]

Answer:

1.129×10⁻⁵ N

1.295 m

Explanation:

Take right to be positive.  Sum of forces on the 31.8 kg mass:

∑F = GM₁m / r₁² − GM₂m / r₂²

∑F = G (M₁ − M₂) m / r²

∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²

∑F = 1.129×10⁻⁵ N

Repeating the same steps, but this time ∑F = 0 and we're solving for r.

∑F = GM₁m / r₁² − GM₂m / r₂²

0 = GM₁m / r₁² − GM₂m / r₂²

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

516 / r² = 207 / (0.482 − r)²

516 (0.482 − r)² = 207 r²

516 (0.232 − 0.964 r + r²) = 207 r²

119.9 − 497.4 r + 516 r² = 207 r²

119.9 − 497.4 r + 309 r² = 0

r = 0.295 or 1.315

r can't be greater than 0.482, so r = 0.295 m.

5 0
3 years ago
The term necrosis means
aksik [14]

Necrosis - the death of most or all of the cells in an organ or tissue due to disease, injury, or failure of the blood supply.

Hope this helps, please mark me brainliest

6 0
3 years ago
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0
3 years ago
What does a trench form?What happens here?
Vinvika [58]
A hole, and people die if they fall in there.
8 0
3 years ago
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