Answer:
3525.19 kg
Explanation:
The computation of the mass of the car is shown below:
As we know that
Fc = m × V^2 ÷ R
m = Fc × R ÷ V^2
Provided that:
Fc = 34.652 kN = 34652 N
R = Radius = 24.98 m
V = speed = 15.67 m/s
So,
m = 34652 × 24.98 ÷ 15.67^2
= 3525.19 kg
Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s
Answer:
125.66 R/s
Explanation:
First 1200 r / min = 20 r/sec
20 r/s * 2pi Radians / r = 40 pi Radians / sec = 125.66 R/s
The distance covered is 1000 m
Explanation:
The rocket is moving by uniformly accelerated motion, so we can find the distance it covers by using the following suvat equation:
where
s is the distance covered
v is the final velocity
t is the time
a is the acceleration
For the rocket in this problem, we have:
v = 445 m/s is the final velocity
is the acceleration
t = 4.50 s is the time
Substituting, we find the distance covered:
Learn more about accelerated motion:
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