Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
<h3>What is electric force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.
Hence the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.
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Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
Answer:
3. they can travel through solids
4. they move rock at right angles to the direction of wave travel
Explanation:
- S waves are called transverse waves they have the ability to move past the solids. They cannot move through the liquids, these waves are perpendicular to the direction of travel.
- They are also called longitudinal waves, the ad is second to record on the seismograph as they slowly pass through the rocks. They have a speed of 3.4 to 7.2 km as per the boundary.
Troposphere, stratosphere, mesosphere, thermosphere, exosphere
Answer:
a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C
Explanation:
Here is the complete question
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?
Solution
a.
i = Q/t = ne/t
n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s
So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C
= 4.98 × 10¹⁹ protons
≅ 5 × 10¹⁹ protons
b
The total kinetic energy of the protons = heat change of target
total kinetic energy of the protons = n × kinetic energy per proton
= 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton
= 30 × 10⁷ J
heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)
ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)
= 30 × 10⁷/14.62
= 2.05 × 10⁷ °C
