Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answers
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
Related Questions
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3 , is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, i
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?
In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.
Tripling the displacement from equilibrium of an object in simple harmonic motion will bring about a change in the magnitude of the object's acceleration by what factor?
Answers
Answer:
56.52 feet³ to the nearest hundredth
Explanation:
the volume of a cone is given as
V =
the radius is 3 feet
height is 6 feet
substituting this given values in the formular
we have that, V = x 3.14 x
x 6
dividing , we have the volume (V)
V= 3.14 x 3 x 6
= 3.14 x 18
= 56.52 feet³ to the nearest hundredth
(B) False
Answers
Answer:
(B) False
Explanation:
No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.
Answers
Answer:
The description is outlined throughout the clarification section following, and according to the given word.
Explanation:
- Throughout the 1960s, Sarnoff Mednick created the RAT as a tool used for testing imaginative convergent thought. Through each RAT test query lists a set of terms, which demands that we have a single additional term that will tie any of the others around.
- Those other words may also be related in something like a variety of ways, such as through creating a compound word or perhaps a semantic connexon.
Answers
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answers
Answer:
69.69 g
Explanation:
Evaporation of water will take out latent heat of vaporization. Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg
Heat taken up by evaporating water
= 2260000 x m J
Heat lost by body
= mass x specific heat of body x drop in temperature
60 x 3500 x .750 ( specific heat of human body is 3.5 kJ/kg.k)
= 157500 J
Heat loss = heat gain
2260000 m= 157500
m = .06969 kg
= 69.69 g
Final answer:
Approximately 78 grams of water would need to evaporate from a 60.0-kg person to lower their body temperature by 0.750ºC. This calculation is based on the principles of thermodynamic heat transfer and the specific body temperature, latent heat of water vaporization, and specific heat capacity of the human body.
Explanation:
To calculate the amount of water mass from an individual's body that would need to evaporate to reduce their body temperature, we can use the principle of thermodynamic heat transfer. The basic equation is Q = mLv, where Q is the heat absorbed or lost, m is the mass, and Lv is the latent heat of vaporization.
In this case, knowing that at body temperature of 37.0°C, the latent heat of water vaporization (Lv) is approximately 2430 kJ/kg, we substitute these numbers. Given our desire to reduce body temperature by 0.750°C in a 60 kg human, we first calculate the amount of heat to dissipate (Q) using Q = mcΔT, where c is the specific heat capacity of the human body (roughly equivalent to that of water, 4.184 kJ/kg°C), m is the mass, and ΔT is the change in temperature.
The calculation is as follows:
Q = (60 kg)(4.184 kJ/kg°C)(0.750°C) = ~189 kJ
Next, we substitute Q into the Q = mLv equation to determine the mass m:
m = Q / Lv = 189 kJ / 2430 kJ/kg = 0.078 kg, or 78 grams
Hence, around 78 grams of water would need to evaporate from a 60.0-kg person to lower their body temperature by 0.750ºC.
Learn more about Heat Transfer here:
#SPJ6
Answers
Answer: 1.11 x 10⁸ Pa
Explanation:
At any deep, the absolute pressure is the same for all points located at the same level, and can be expressed as follows:
p = p₀ + δ. g . h, where p₀ = atmospheric pressure = 101, 325 Pa
Replacing by the values, we get:
p= 101,325 Pa + 1025 Kg/m³ . 9.8 m/s². 11,033 m = 1.11 x 10⁸ Pa.