A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case
1 answer:
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The answer is D. 32 m.
The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
⇒ 
It is given:
t = 10 s
d = ?
So:
The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
It is given:
t = 10 s
d = ?
So:
Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity
The initial velocity is the horizontal component
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = t +
gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² ,
Substitute these values in the rule
→ -60 = 0 + (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* <em>The ball is in the air for 3.5 seconds </em>
The initial velocity is the horizontal component
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = (3.5)
Divide both sides by 3.5
→ = 28.57 m/s
<em>The initial horizontal component of velocity is 28.6 m/s</em>
The vertical component of the final velocity is
→ =
+ gt
→ = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ = 0 + (-9.8)(3.5)
→ = -34.3 m/s
<em>The vertical component of the final velocity is 34.3 m/s downward</em>
The final velocity v is the resultant vector of and
→ Its magnetude is
→ Its direction
→ = 28.6 ,
= -34.3
Substitute this values in the rules above
→
→ Its direction
The negative sign means the direction is below the horizontal
<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>
Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:
Given mass of box = 5.0 kg
Frictional force = 30 N
thus
Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero
Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.