_________________________________
If west means the west of the axis x the velocity equal :
(a) The spring stiffness constant of the spring is 18,392 N/m.
(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.
<h3>Kinetic energy of the car</h3>
The kinetic energy of the car is calculated as follows;
K.E = ¹/₂mv²
K.E = ¹/₂ x 950 x 22²
K.E = 229,900 J
<h3>Stiffness constant of the spring</h3>
The stiffness constant of the spring is calculated as follows;
K.E = U = ¹/₂kx²
k = 2U/x²
k = (2 x 229,900)/(5)²
k = 18,392 N/m
<h3>Force exerted on the spring</h3>
F = kx
F = 18,392 x 5
F = 91,960 N
<h3>Time of impact</h3>
F = mv/t
t = mv/F
t = (950 x 22)/(91960)
t = 0.23 s
Learn more about spring constant here: brainly.com/question/1968517
#SPJ4
1) metal
Even though metalloids are also conductors of heat and electricity, malleable they are not as good as metals.
Metals are very good conductors of electricity and heat. They are also very hard to touch. Noble gases and non metals are the exact opposite in physical and chemical properties. Metals readily react with oxygen.
Answer:
Rebounce angle is 345°
Rebounce speed is 989.95m/s
Explanation:
Calculate the x component of the velocity of the bullet before impact by using the following relation:
Vbx= Vb Cos thetha
Here, is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°
Substituting
Vbx = Cos15 ×1400 = 1352.30m/s
Calculate the y component using the relation:
Vby = Vo Sin theta
Vby = sin 15° × 1400
Vby = 362.35m/s
The rebounce angle = 360 - incidence angle
Rebounce angle =( 360 - 15)° = 345°
The rebound speed V' = Vby - Vbx
V' = (1352.30 - 362.35)m/s
V' = 989.95 m/s
The answer is near the poles.
