Answer:
Acceleration of the car will be 
Explanation:
We have given that car starts from rest so initial velocity of the car u = 0 m/sec
And car traveled 400 m in 10 sec
So distance traveled by car s = 400 m
Time taken to compete this distance t = 10 sec
We have to find the acceleration of the car
From second equation of motion we know that 
So 

So acceleration of the car will be 
The answer would be, the iris. Hope this helps:)
Answer:
L = 0.475 m = 475 mm = 18.7 inches
Explanation:
A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lb ) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in).
E = 207 GPa = 207*10⁹ Pa
D = 10.2 mm = 0.0102 m
P = 8900 N
ΔL = 0.25 mm = 2.5*10⁻⁴ m
L = ?
We can use the Equation of the Hooke's Law
ΔL = P*L / (A*E) ⇒ L = ΔL*A*E / P
⇒ L = (2.5*10⁻⁴ m)*(π*(0.0102 m)²*0.25)*(207*10⁹ Pa) / (8900 N)
⇒ L = 0.475 m = 475 mm = 18.7 inches
Answer:
Explanation:
Given
Acceleration a = 1.0m/s²
Displacement S = 1.0m
Required
Time t taken by the leaf to displace
Using the equation of motion
S = ut+1/2at²
Substitute
1.0 = 0+1/2(1)t²
1 = t²/2
Cross multiply
t² = 2
t = ±√2
t = 1.41secs
It takes the leaf to 1.41s to displace by 1m upward