Answer:
The atomic bomb, is as defined by britannica.com “a deadly weapon caused by the sudden release of energy after the splitting, or fission, of the nuclei of heavy elements like uranium.” In 1945, the United States (US) dropped two atomic bombs, one in Hiroshima and the other in Nagasaki ending WWII.
Explanation:
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Draw a free body diagram to show which forces act in the x and y directions. The x component equation is σfx = 0. The σfx being all the forces acting in the x direction.
Answer:
Option D. ²³⁹₉₃Np
Explanation:
Let the unknown be ʸₓA.
Thus, the equation becomes:
²³⁹₉₂U —> ⁰₋₁e + ʸₓA
Next, we shall determine the x, y and A. This can be obtained as follow:
92 = –1 + x
Collect like terms
92 + 1 = x
93 = x
x = 93
239 = 0 + y
239 = y
y = 239
ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np
Thus, the complete equation is:
²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np
The most massivest stars end their lives as black holes. <em>(D)</em>
(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671