The force of gravity produces acceleration in all C. freely falling objects and this is known as acceleration due to gravity
Explanation:
A body is said to be in free fall when there is only one force acting on the body: the force of gravity.
Gravity is a force that acts downward, i.e. towards the Earth's centre.
If we are near the Earth's surface, the magnitude of the force of gravity on a body is given by
where:
m is the mass of the body
g is known as the acceleration of gravity , whose value near the Earth's surface is 
).
We can apply Newton's second law on an object in free-fall, to find its acceleration. In fact, we have:
where F is the force acting on the body and a is its acceleration.
Solving for the acceleration,
And substituting F,
Therefore, every object in free-fall accelerates at towards the ground.
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we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m
Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)
Therefore
Ui = mgL(1-cos45)
Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)
Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2
According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s
<span>The speed f T after swing is 7.89 m/s</span>
Explanation:
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