Answer:
Minimum stopping distance = 164.69 ft
Explanation:
Speed of car = 35.6 mi/hr = 15.82 m/s
Stopping distance = 40.8 ft = 12.44 m
We have equation of motion
v² = u² + 2as
0²=15.82²+ 2 x a x 12.44
a = -10.06 m/s²
Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.
Speed of car = 71.5 mi/h = 31.78 m/s
We have
v² = u² + 2as
0² = 31.78² - 2 x 10.06 x s
s = 50.2 m = 164.69 ft
Minimum stopping distance = 164.69 ft
Answer:
16J
Explanation:
From hookes law
The work done in a spring is given as W =1/2ke^2
Given that the force constant (k) is constant in the spring material
We have that 2W = e^2
Let W1 = 4J e1= 2cm e2 = 4cm
Let W2 be the work required to stretch it an additional 4cm
W1/ W2 = e1^2/e2^2
W2 = W1* e2^2 / e1^2
= 4* 4^2 /2^2
=4× 16 / 4
= 16J
Answer: momentum has the same direction as that of velocity but when 2 bodies with the same linear momentum & different velocities it has different masses because a vector quantity is represented by a cross product of mass and velocity of object .
D. 1609 meters
5280/3.281 = 1609m