Initial velocity, u = 4 m/s
acceleration due to gusts of wind = 3 m/s^2
time, t = 1 min = 60 s
Let distance travelled = S
From equation of motion,

Thus, the boat would have traveled 5640m after gusts picked up.
There's supposed to be a blank in the statement, where the answer is supposed to be inserted.
The question is supposed to say: "A wave is a disturbance that carries ______ from one place to another through matter or space".
To answer the question, write the word "<em>energy</em>" in the blank.
Answer:
Explanation:
Given
speed of Electron 
final speed of Electron 
distance traveled 
using equation of motion

where v=Final velocity
u=initial velocity
a=acceleration
s=displacement


acceleration is given by 
where q=charge of electron
m=mass of electron
E=electric Field strength

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:

Explanation:
Ok, the average speed can be calculate with the next equation:
(1)
Basically the car cover the same distance "d" two times, but at different speeds, so:

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

From basic physics we know:

so:


Using the previous information in equation (1)

Factoring:
(2)
Finally, replacing the data in (2)
