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
Hi!Schrodinger equation is written as HΨ = EΨ, where h is said to be a Hamiltonian operator.
What What What What...... error
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament, 
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
Answer:
<u>8.13 kJ</u>
Explanation:
Q = mCT where Q is the energy (in Joules, here), m is the mass (kg, here) C is the specific heat, and T is the temperature (C).
Q = mCT
Q = (1.5kg)*((387J/(kg*C))*(14C)
Q = 8127 J, or 8.13 kJ with 3 sig figs.
