:<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>
<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>
Explanation:
It is given that,
Mass of Madeleine, 
Initial speed of Madeleine, 
(due east)
Final speed of Madeleine, 
(due west)
Mass of Buffy, 
Final speed of Buffy, 
(due east)
Let 
is the Buffy's velocity just before the collision. Using the conservation of linear momentum as :
So, the initial speed of the Buffy just before the collision is 7.13 m/s and it is moving due west. Hence, this is the required solution.
I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?
Answer:
<h3>0.145m</h3>
Explanation:
Using the equation of motion formula y = ut + 1/2gt² where;
y is the horizontal displacement
u is the initial velocity
g is the acceleration due to gravity
t is the time
To calculate the horizontal displacement, we need to first get the time t. Using the equation:
v = u+gt
1.3² = 0+(9.8)t
1.69 = 9.8t
t = 1.69/9.8
t = 0.172s
Substituting the time into the equation above to get y
y = 0+1/2(9.8)(0.172)²
y = 4.905(0.029584)
y = 0.145m
Hence the cup's horizontal displacement during the fall is 0.145m
