The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus,
1.22 a = 12.0 - T (eqn 1)
and for the 20.0 N block:
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction)
2.04 a = T - 6.5 (eqn 2)
[eqn 1] + [eqn 2] → 3.26 a = 5.5
a = 1.69 m/s²
Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5
T = 9.95 N
Now want the resultant force acting on the 20.0 N block:
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N
<span>Units have to be consistent ... so have to convert 75.0 cm to m: </span>
75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m
<span>work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J</span>
Answer: yes
Step-by-step explanation:
Answer:
$31728
Step-by-step explanation:
sorry if I'm wrong
-a²-3b³+c²+2b³-c²
= -(3)² - (3×2)³ + (-3)² + (2 × 2)³ - ( -3)²
= 9 - 24 + 9 + 16 - 9
= 1
answer is 1.
Answer:
The position P is:
ft <u><em> Remember that the position is a vector. Observe the attached image</em></u>
Step-by-step explanation:
The equation that describes the height as a function of time of an object that moves in a parabolic trajectory with an initial velocity 
is:
Where 
is the initial height = 0 for this case
We know that the initial velocity is:
82 ft/sec at an angle of 58 ° with respect to the ground.
So:
ft/sec
ft/sec
Thus
The height after 2 sec is:
Then the equation that describes the horizontal position of the ball is
Where
for this case
ft / sec
ft/sec
So
After 2 seconds the horizontal distance reached by the ball is:
Finally the vector position P is:
ft
