Answer:
(c). The two blocks end in a tie
Explanation:
the reason being the absence of any resistance offered to both of the blocks.
if the slope of the hill is for instance 60 deg.
then the acceleration in absence of any resistance is a= 9.81sin(60)
since the acceleration is same then both of the blocks will reach at the same instant
The time difference between their landing is 2.04 seconds.
<h3>
Time of difference of the two balls</h3>
The ball thrown vertical upwards will take double of the time taken by the ball thrown vertically downwards.
Time difference, = 2t - t = t
t = √(2h/g)
where;
- h is the height of fall
- g is acceleration due to gravity
Apply the principle of conservation of energy;
¹/₂mv² = mgh
h = v²/2g
where;
h = (20²)/(2 x 9.8)
h = 20.41 m
<h3>Time of motion</h3>
t = √(2 x 20.41 / 9.8)
t = 2.04 s
Thus, the time difference between their landing is 2.04 seconds.
Learn more about time of motion here: brainly.com/question/2364404
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Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
Answer:
163.8 ft
Explanation:
In triangle ABD
= 155 ft
Using Pythagorean theorem in triangle ADC
= distance between the anchor points
distance between the anchor points is given as
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
