Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero. So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".
If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.
And if there are four ropes holding it up, then the sum of
the four tensions is 100N. If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.
Correct answer choice is:
D. A continuous transmission of energy from one location to the next.
Explanation:
Waves include the carrier of energy without the carrier of matter. In outcome, a wave can be characterized as a change that progresses into a medium, carrying energy from one spot (its source) to different spot without carrying matter.
Answer:
B
Explanation:
this is because the neutrons do not have a charge, the things that have charge in an atom are electrons and protons.
and in the nucleus of an atom, there are protons and neutrons so you can see that A is not the answer
if you see the periodic table, you will know that the number of electrons and protons are equal, so the charges cancel each other out, hence the charge of an atom will be neutral
let me give you a tip which I got from my teacher, never write there is no charge in the atom, this suggests that there is no protons or electrons.
instead, write, the it is neutral
hope it helps if not please report it so that someone else gets to try it out
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
