Answer:
KE = 1/2 M v^2 kinetic energy of projectile
KE = 1/2 * 20 * (54 m/s)^2 = work done on projectile
W = 10 * 54^2 = work done on projectile
W = 29,160 J
Answer:
Explanation:
<u>Dynamics</u>
When a particle of mass m is subject to a net force F, it moves at an acceleration given by
The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by
The variable acceleration is calculated by:
The instant velocity is the integral of the acceleration:
Integrating
<u>Answer:</u>
I think its
<em>10 kg</em>
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Answer: The correct answer is- Field.
Explanation -
The modified space around a charged particle is known as electric field.
The strength of electric field at a point is calculated as the force experienced by a unit test charge present at that point.
Thus, a region of space in which a test charge experiences a force is referred to as Field.
Answer:
Stress = F / A force per unit area
A = 3.00 cm^2 = 3 E-4 m^2
F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied
F/3 = 2.4E4 N if force not to exceed limit (= f)
f = M a
a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g
