:<span> </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s
At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)
So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..
Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.
ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²
►v = √198 = 14.0 m/s</span>
(a) 
The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

where
I is the intensity of the wave
c is the speed of light
In this problem,

and substituting
, we find the radiation pressure

(b) 
Since we know the cross-sectional area of the laser beam:

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

And then, since we know the mass of the atom

we can find the acceleration, by using Newton's second law:

Centripetal force <span>a force that acts on a body moving in a circular path and is directed toward the center around which the body is moving. It is calculated by the expression:
F = mv^2/r
where m is the mass, v is the velocity and r is the radius.
F = 7.26(31.95)^2 / (1.215) = 6100 N</span>
Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

hence, B_o = 1.013μT
Color <span>is a physical property of all visible light determined by the light's frequency and visible to the human eye.</span>