Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Answer:
404K
Explanation:
Data given, Kinetic Energy.K.E=8.37*10^-21J
Note: as the temperature of a is increase, the rate of random movement will increase, hence leading to more collision per unit time. Hence we can say that the relationship between the kinetic energy and the temperature is a direct variation.
This relationship can be expressed as
where K is a constant of value 1.38*10^-23
Hence if we substitute the values, we arrive at
converting to degree we have 
False, its up to you to decide what you belive is right and what is wrong.
Answer:
The speed of the clay before the impact was 106.35 m/s.
Explanation:
the only force doing work on the system is the frictional force, f, the work done by f is given by:
Wf = ΔK = Kf - Ki
The clay and the block will come to rest after sliding Δx = 7.50 m, if their intial speed is v and the combined mass is m and μ is the coefficient of friction and g is gravity,then:
f×Δx = Ki
m×g×Δx×μ = 1/2×m×v^2
v^2 = 2×g×Δx×μ
= 2×(9.8)×(7.50)×(0.650)
= 95.55
v = 9.78 m/s
This is the veloty of clay and block after the clay hit the block.
if the velocity the clay and block attains after the impact is v and the initial speed of the clay is v1 and the mass is m and the speed of the block initially is V = 0 m/s and the mass is M, then according to the conservation of linear momentum:
m×v1 +M×V = v(m + M)
m×v1 = v(m + M)
v1 = v(m + M)/m
v1 = (9.78)(8.3×10^-3 + 82×10^-3)/(8.3×10^-3)
v1 = 106.35 m/s
Therefore, the speed of the clay before the impact was 106.35 m/s.
