Answer:
It is easier to hear a musician in the classroom than outdoors
Explanation:
It is easier to hear a musician in the classroom due to the improved acoustics provided by the walls of the classroom whereby along with the direct sound of the musician, which is the lead source of the sounds, there is an increased number of indirect sound reaching the ear in the classroom than outdoors and due to precedence effect, all the sound appear to come from the musician
In music played outside, along side the direct sound from the musician, the indirect sound that reach the ear is echoed from maybe by only the ground while the majority of the sound from the music wanders away with the wind and in other directions as well as being absorbed such that speakers will be required to improve the sound of the music outdoors.
Answer:
Explanation:
Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².
Now, the pilot is traveling in a circle of radius
r = 3340 m
And the speed is
v = 495 m/s
Then, acceleration?
The acceleration of a circular motion can be determine using centripetal acceleration
a = v² / r
a = 495² / 3340
a = 73.36 m/s².
Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully
Answer:
Your pinball machine was built using two kinds of simple machines: a lever and an inclined plane. The lever shot the marble to the top of the box with lots of force. The inclined planes made the marble wind its way down to the bottom.
Answer:
Magnitude of static friction force is 70 sin40° = 44.99 N.
No, it is not necessary that it is maximum static friction.
Normal force is equal to 70 cos40° = 53.62 N.
Explanation:
We apply newton law of moton equation along the plane and perpendicular to plane;
Along the plane,
70 sin 40° = 
---------------(1)
70 cos 40° = N --------------(2)
= μN -----------------(3)
So, it depends on the value of μ that the friction is maximum or not .
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.
The work will be completely defined by
(Force) x (distance in the y-direction),
and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.
We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.
From y=0 to y=2.40 is a distance of 2.40 upward.
Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.
Work = integral of (F·dy) evaluated from 0 to 2.40
= integral of (-2.85 y² dy) evaluated from 0 to 2.40
= (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .
Now, integral of (y² dy) = 1/3 y³ .
Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)
= 1/3 · 13.824 = 4.608 .
And the work = (-2.85) · the integral
= (-2.85) · (4.608)
= - 13.133 .
-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.
-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.
-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.
As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
