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MariettaO [177]
3 years ago
5

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Physics
1 answer:
Pachacha [2.7K]3 years ago
6 0

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

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As moleclues get further apart, i.e. As a substance changes state from solid to liquid to gas, molecules gain kinetic energy and vibrate/move more. This means they gain heat energy (the averge energy a substance has) so the temperature increases

Substances exist in different states at different temperatures and different substances will exist in different states at the same temperature. This is to do with the forces between molecules and how much heat (energy) is required to break them
5 0
3 years ago
A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup
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Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) =  7.5 mol

As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)

When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V= \frac{(7.5 \, mol)*(8.314 \, J/(mol-K))*(300 \, K)}{(92205.75 \, Pa)} =0.2029 \, m^{3}

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3 0
3 years ago
The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi
vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

Q=cm\Delta T (1)

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\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}

\Delta T=-\frac{Q}{16744}

Sample B:

\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}

\Delta T=-\frac{Q}{5400}

Sample C:

\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}

\Delta T=-\frac{Q}{9450}

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression -\frac{Q}{cm} that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of \Delta T implies the temperature is dropping)

8 0
3 years ago
a wave is described by where x is in meters, y is in centimeters and t is in seconds. The angular frequency is
Sergeeva-Olga [200]

Complete Question

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave  frequency is

Answer:

The  value is w =  10 \ rad /s

Explanation:

From the question we are told that  

    The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)

Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

         y(x,t) =  Asin(kx + wt )

Where  w  is the  angular frequency

Now comparing this equation  with that given we see that

       w =  10 \ rad /s

 

               

7 0
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Answer:

576 joules

Explanation:

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radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

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K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
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