:<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>
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components


Acceleration components




Time taken is 0.26088 seconds

The distance the beetle travels on the ground is 0.3677181864 m
I believe the answer is 2m/s
Answer:
height is 69.68 m
Explanation:
given data
before it hits the ground = 46 % of entire distance
to find out
the height
solution
we know here acceleration and displacement that is
d = (0.5)gt² ..............1
here d is distance and g is acceleration and t is time
so when object falling it will be
h = 4.9 t² ....................2
and in 1st part of question
we have (100% - 46% ) = 54 %
so falling objects will be there
0.54 h = 4.9 (t-1)² ...................3
so
now we have 2 equation with unknown
we equate both equation
1st equation already solve for h
substitute h in the second equation and find t
0.54 × 4.9 t² = 4.9 (t-1)²
t = 0.576 s and 3.771 s
we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls
so take t = 3.771 s
then h from equation 2
h = 4.9 t²
h = 4.9 (3.771)²
h = 69.68 m
so height is 69.68 m
Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.