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3 years ago

What would the electrostatic force be for two 0.005C charges 50m apart?

Physics

1 answer:

irakobra [83]3 years ago

3 0

Answer:

90 N

Explanation:

The electrostatic force between two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem we have

q1 = q2 = 0.005 C

r = 50 m

So the electrostatic force is

$F=(9\cdot 10^9 N m^2 C^{-2})\frac{(0.005 C)^2}{(50 m)^2}=90 N$

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you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made

Nuetrik [128]

You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.

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3 years ago

What two types of scientific knowledge can be expressed as mathematical equations? (Please don't tell me the answer could you pl

gtnhenbr [62]

An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.

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3 years ago

three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25

ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0

3 years ago

A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K).

mina [271]

Answer:

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

$P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa$

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

$P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa$

(c) Same steps as in part (a)

$P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa$

ΔP=9.164-9.14

ΔP=0.0245kPa

8 0

3 years ago

A vertical spring stretches 3.4 cm when a 12-g object is hung from it. The object is replaced with a block of mass 26 g that osc

Mariulka [41]

Answer:

Period of motion is approximately 0.5447 seconds

Explanation:

We start by calculating the constant "k" of the spring which can be derived from the fact that an object of mass 12 g produced a stretch of 3.4 cm: (we write everything in SI units)

F = k * x

0.012 kg * 9.8 m/s^2 = k 0.034 m

k = 0.012 kg * 9.8 m/s^2 / (0.034 m)

k = 3.46 N/m

now we use the formula for the period (T) of a spring of constant k with a hanging mass 'm':

$T=2\pi\,\sqrt{\frac{m}{k} }$

which in our case becomes:

$T=2\pi\,\sqrt{\frac{0.026}{3.46} } \approx 0.5447\,\,sec$

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3 years ago