The kinetic energy in the first case is 4 times more than the second case.
Hence, option D)It is 4 times greater is the correct answer.
<h3>What is Kinetic Energy?</h3>
Kinetic energy is simply a form of energy a particle or object possesses due to its motion.
It is expressed as;
K = (1/2)mv²
Where m is mass of the object and v is its velocity.
Given that;
- For the first case, velocity v = 16m/s
- For the second case, velocity = 8m/s
- Let the mass of the car be m
For the first case, kinetic energy of the car will be;
K = (1/2)mv²
K = (1/2) × m × (16m/s)²
K = (1/2) × m × 256m²/s²
K = mass × 128m²/s²
For the second case, kinetic energy of the car will be;
K = (1/2)mv²
K = (1/2) × m × (8m/s)²
K = (1/2) × m × 64m²/s²
K = mass × 32m²/s²
Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.
Hence, option D)It is 4 times greater is the correct answer.
Learn more about kinetic energy here: brainly.com/question/12669551
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Answer:
The work done is 205 kJ.
Explanation:
Hi there!
Work can be calculated using the following equation:
W = F · Δx
Where:
W = work
F = applied force
Δx = displacement
In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:
W = ∫ F · dx
F = 3.6 N/m³ · x³ - 76 N
W = ∫ (3.6 x³ - 76)dx
W = 0.9 x⁴ - 76x
Evaluating from xi to xf:
W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m
W = 205 kJ
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
The awnser would be True. Hope I helped!
