Answer:
I think the answer is C/ Tell me If you got it right
v₀ = initial velocity of the mobile = 10 m/s
v = final velocity of the mobile = 20 m/s
a = acceleration of the mobile = 5 m/s²
d = distance traveled during this operation = ?
Using the kinematics equation
v² = v²₀ + 2 a d
inserting the above values in the equation
20² = 10² + 2 (5) d
400 = 100 + 10 d
subtracting 100 both side
400 - 100 = 100 - 100 + 10 d
300 = 10 d
dividing both side by 10
300/10 = 10 d/10
d = 30 m
hence mobile travels 30 m.
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m
Answer:
Can't see anything, please share clearly
