1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of
θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.
(a) Find the value of the normal force on the block.
(b) Find the magnitude of force F~2 that is acting on the block
(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .
(a) Find the value of the normal force on the block.
(b) Find the magnitude of force F~2 that is acting on the block
(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .
1 answer:
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Answer:
"0.758315 sec" is the appropriate choice.
Explanation:
The given values are:
Angular speed,
= 29 rad/s
or,
=
Tire rotates,
= 3.5 times
Now,
The time interval will be:
⇒
⇒
⇒
⇒
Answer:
5.2m/s^2
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve
for this case we can use the ecuation number 3
x=100m
t=6.2s
Vo=0m/s