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sergeinik [125]
3 years ago
13

How can a nuke (Nuclear bomb) be hotter than the sun?

Physics
1 answer:
Art [367]3 years ago
6 0
Cause it can take so much heat then when the sun beams down on it then the process gets alot more complicated it gets a little hotter than it starts to get more and more hotter .
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A 0.2 kg cannon ball is fired at an upward angle of 45° from the top of a 165 m cliff with a speed of 175 m/s. (A) Using conserv
sergejj [24]

To solve this problem, it is necessary to apply the concepts related to the work done by a body when a certain distance is displaced and the conservation of energy when it is consumed in kinetic and potential energy mode in the final and initial state. The energy conservation equation is given by:

\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f

Where,

KE = Kinetic Energy (Initial and Final)

PE = Potential Energy (Initial and Final)

And the other hand we have the Work energy theorem given by

\Delta KE = W = F*d

Where

W= Work

F = Force

D = displacement,

PART A) Using the conservation of momentum we  can find the speed, so

\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f

\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+mg_h2

The height at the end is 0m. Then replacing our values

\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+0

Deleting the mass in both sides,

\frac{1}{2}v_1^2 + gh_i = \frac{1}{2}v_f^2

Re-arrange for find v_f^2,

v_f^2= 2(gh_i)+v_1^2

v_f^2 = 2(9.8*165)+(175)^2

v_f=\sqrt{33859}

v_f = 184.008m/s

PART B) Applying the previous  Energy Theorem,

\Delta KE = W = F*d

\frac{1}{2}mv^2 = F*d

\frac{1}{2}(0.2)(184.008)^2 = (75)*d

Solving for d

d = 45.15 m

4 0
3 years ago
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