Answer:
hi. :)))))))))))))))))))))
Answer:
a.)1.)12m/s
2.)12m/s
3.)7m/s
4.)7m/s
5.)14m/s
6.)0m/s
b.)1.) 3m/s^2
2.)1.71m/s^2
3.).58m/s^2
4.).39m/s^2
5.).70m/s^2
6.)0
c.)1.)48m
2.)84m
3.)84m
4.)126m
5.)280m
6.)suma todos los metros mas 98m y obtendras la distancia en cual el carro se para
Explanation:
vinicial=0m/s
tiempo inicial=0s
Intervalo 1
v=12m/s
t=4s
aceleracion= vf-vi/t
=12-0/4
=12/4=3m/s^2
Intervalo 2
v=12m/s
t=7s
a=12/7
Intervalo 3
v=7m/s
t=12s
a=7/12
Intervalo 4
v=7m/s
t=18s
a=7/18
Intervalo 5
v=14m/s
t=20s
a=14/20
Intervalo 6
v=0m/s
t=27s
a=0/27
Answer:
Stress = F / A force per unit area
A = 3.00 cm^2 = 3 E-4 m^2
F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied
F/3 = 2.4E4 N if force not to exceed limit (= f)
f = M a
a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g
Answer:
Her angular velocity when tucked is greater than when straight by a factor of 0.23
Explanation:
Moment of inertia (I) = mr^2 = mv^2/w^2
m is mass of the diver
v is diver's linear velocity
w is her angular velocity
When straight, I = 14 kg.m^2
mv^2/w^2 = 14
w^2 = mv^2/14
w = sqrt(mv^2/14) = 0.27sqrt(mv^2)
When tucked, I = 4 kg.m^2
w^2 = mv^2/4
w = sqrt(mv^2/4) = 0.5sqrt(mv^2)
Her angular velocity when tucked is greater than when straight by 0.23 (0.5 - 0.27 = 0.23)
