Answer:
True
Explanation:
A crowbar makes our work easier by multiply effort because it belongs to first class lever.
And first class lever makes work easier by multiplying the effort
Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
The range of the projectile is 188 m
Explanation:
The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
The path of a projectile is the combination of these two motions: see figure in attachment.
In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.
We have:
t = 5.0 s (time of fligth of the projectile)
and the horizontal velocity is constant, and it is given by

where
is the initial velocity
is the angle of projection
Substituting,

And therefore, the range of the projectile is:

Learn more about projectile motion:
brainly.com/question/8751410
#LearnwithBrainly
If the object is kept in between the principle axis and the focus but some what nearer to the focus then we will get the enlarge,erect,and real image.