The centripetal force is:
F = mv² / R
Where:
m: mass of the object
v: object speed
R: radius of the curve.
We have to:
m = 2000kg
v = 25 m / s
R = 80 meters.
Then the centripetal force acting on the vehicle is:
F = (2000kg * (25m / s) ²) / 80m
F = 15625 N
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Answer:
D
Explanation:
First we define our variables
V0=29.4
a=-9.8
V=0
We have to find the maximum displacement , which I will define as X
We use formula v^2=v0^2+2aX
All we do is substitute our values
0=29.4^2-19.6X
29.4^2=19.6X
X=29.4^2/19.6=44.1
Answer:
t = 3/8 seconds
Explanation:
h=-16t^2 - 10t+6
h= 0 when it hits the ground
0=-16t^2 - 10t+6
factor out a -2
0= -2(8t^2 +5t -3)
divide by -2
0 = (8t^2 +5t -3)
factor
0=(8t-3) (t+1)
using the zero product property
8t-3 = 0 t+1 =0
8t = 3 t= -1
t = 3/8 t= -1
t cannot be negative ( no negative time)
t = 3/8 seconds
Answer:
The answer is 4 pounds
Explanation:
The explanation is that 1 kilogram is equal to 2 pounds so multiply the kilogram with the 1 pound