Answer:
The value of tangential acceleration 
40 
The value of radial acceleration 
Explanation:
Angular acceleration = 50 
Radius of the disk = 0.8 m
Angular velocity = 10 
We know that tangential acceleration is given by the formula 
Where r = radius of the disk
= angular acceleration
⇒ 0.8 × 50
⇒ 40
This is the value of tangential acceleration.
Radial acceleration is given by
Where V = velocity of the disk = r 
⇒ V = 0.8 × 10
⇒ V = 8 
Radial acceleration
This is the value of radial acceleration.
Answer:
A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.
A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.
An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.
I HOPE THIS IS HELPFUL.
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Answer:
Explanation:
In the x direction the force will be
½(-w₀)L/2 = -¼w₀L
acting ⅔(L/2) = L/3 below the x axis.
In the y direction the force will be
½(-w₀)L + ½w₀L/2 = -¼w₀L
the magnitude of the resultant will be
F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛
in the direction
θ = arctan(-¼w₀L / -¼w₀L) = 225°
to find the distance, we balance moments
(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]
(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]
(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]
(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L
(√⅛)[d] = 5L/24
d = 5L/24 / (√⅛)
d = 5√⅛L/3
Answer: a) 11.76 m/s b) 7.056 m
Explanation:
The described situation is as follows:
An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.
This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:
(1)
(2)
Where:
Is the final velocity of the object
Is the initial velocity of the object (it was dropped)
is the acceleration due gravity
is the height of the tower
is the time it takes to the object to reach the ground
b) Begining with (1):
(3)
(4)
(5) This is the final velocity of the object
a) Substituting (5) in (2):
(6)
Clearing :
(7)
(8) This is the height of the tower
