1. Strzała o masie 20 g tuż po wystrzale ma prędkość 50 m/s. Oblicz pracę wykonaną

przez zawodniczkę. Ile wynosi energia potencjalna sprężystości napiętej cięciwy?

2. Na trampolinę o współczynniku sprężystości k=2 kN/m stanął zawodnik o masie 80 kg. O ile obniży się krawędź trampoliny?

Explanation:

Question 1

An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?

mass m = 20g = 20/1000 = 0.02kg

speed v = 50m / s

P.E = K.E = ½mv²

P.E = ½ × 0.02 × 50²

P.E = 25 J

work done = P.E = 25J

Qestion 2

A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.

force of elasticity

F = -kx

x = F / k

in our case F will be the force of pressure or gravity

F = mg

g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg

F = 80kg * 10N / kg = 800 N

x = 800N / -2000N = -0.4

The trampoline will lower, so from the level by 0.4 meters and hence this minus

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.