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3 years ago

How is the electrostatic force affected when the magnitude of a charge is doubled?

1 answer:

4 0

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>

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In and electric circuit, where do the electrons come from that flow in the circuit

Firdavs [7]

The electrons are already there. They are freely moving through the conductor.

7 0

2 years ago

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A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to

uysha [10]

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

6 0

2 years ago

Devise an exponential decay function that fits the given data, then answer the accompanying questions. Be sure to identify the

7nadin3 [17]

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

6 0

2 years ago

A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observ

siniylev [52]

Answer:

$m=1.53kg$

Explanation:

To solve this problem we use the Hooke's Law:

$F=k*\Delta x$ (1)

F is the Force needed to expand or compress the spring by a distance Δx.

The spring stretches 0.2cm per Newton, in other words:

1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm

The force applied is due to the weight

$F=mg$

We replace in (1):

$mg=k*\Delta x$

We solve the equation for m:

$m=k*\Delta x/g=5*3/9.81=1.53kg$

7 0

2 years ago

Which change will always result in an increase in the gravitational force between 2 objects

Nat2105 [25]

<span>Reducing the distance between them. In theory, also increasing the mass; but you can't really change the mass of an object. However, you can compare the forces if you replace an object by a different object, which has a different mass.
</span>
i hope this will work..

4 0

3 years ago

Read 2 more answers