Answer:
U= 4.1×10^{-3} J
Explanation:
Given:
Young's modulus E= 2.5×10^5 N/m^2
thickness of material L_o =3.0 mm
Area A= 0.50 cm^2
ΔL= 1.0 cm
The potential energy is given by
U= 1/2FΔx
therefore
now putting values we get
U= 4.1×10^{-3} J
Answer:
The spring constant required = 949.2 N/m
Explanation:
The energy stored in the spring is converted to kinetic energy of the projectiles fired.
Energy stored in spring's = kx²/2
Kinetic energy of bullets = mv²/2
k = spring constant = ?
x = maximum compression of the spring = 8 cm = 0.08 m
m = mass of projectiles fired = 3.0 g = 0.003 kg
v = velocity of projectiles fired = 45.0 m/s
kx² = mv²
k (0.08²) = 0.003 (45²) = 6.075
k = 949.2 N/m
The spring constant required = 949.2 N/m
Complete question
A cannon with a muzzle velocity of 500. meters per second fires a cannonball at an angle of 30.° above the horizontal. What is the vertical component of the cannonball's velocity as it leaves the cannon?
A 0.0 m/s
B 250. m/s
C 433 m/s
D 500. m/s
Answer:
The correct option is C
Explanation:
From the question we are told that
The velocity is
The angle is
Generally the vertical component of the canon ball is mathematically represented as
=>
=>
Answer:
Explanation:
As we know that we board in the car of ferris wheel at the bottom position
So we will have
final height of the car at angular displacement given as
here we know that
so we have