Speed = distance/time
speed= 122÷27=4.52m/s (3sf)
Answer:
Explanation:
First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volume in liters, 'n' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmospheres per moles Kelvin).
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
Answer:
Fy=107.2 N
Explanation:
Conceptual analysis
For a right triangle :
sinβ = y/h formula (1)
cosβ = x/h formula (2)
x: side adjacent to the β angle
y: opposite side of the β angle
h: hypotenuse
Known data
h = T = 153.8 N : rope tension
β= 44.2°with the horizontal (x)
Problem development
We apply the formula (1) to calculate Ty : vertical component of the rope force.
sin44.2° = Ty/153.8 N
Ty = (153.8 N ) *(sen44.2°)= 107.2 N directed down
for equilibrium system
Fy= Ty=107.2 N
Fy=107.2 N upward component of the force acting on the stake