We use the formula,
.
Here, v is velocity and its value given 26 mi/h ( in m/s,
) and d is distance and its value is given 80 m.
Substituting these values in above formula we get,

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s
Answer:
kinetic friction may be greater than 400 N or smaller than 400 N
Explanation:
As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

now we know that

so here value of limiting static friction force is always more than kinetic friction
also we know that
initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction
and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction
so kinetic friction may be greater than 400 N or smaller than 400 N
Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Answer:
Explanation:
Volume of the insulating shell is,

Charge density of the shell is,

Here, 

B)
The electric field is 
For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.
C)
For R <r <2R According to gauss law

substitute 

D)
The net charge enclosed for each r in this range is positive and the electric field is outward
E)
For r>2R
Charge enclosed is zero, so electric field is zero
To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m