Answer: Temperature and humidity are the two characteristics used to classify air masses.
Explanation:
His total displacement from his original position is -1 m
We know that total displacement of an object from a position x to a position x', d = final position - initial position.
d = x' - x
If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.
Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.
His final position, x" after moving back 6 m is gotten from
x" - x' = -6 m
x" = -6 + x'
x" = -6 + 5
x" = -1 m
Thus, his total displacement from his original position is
d = final position - initial position
d = x" - x
d = -1 m - 0 m
d = -1 m
So, his total displacement from his original position is -1 m
Learn more about displacement here:
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Answer:
11.7 m/s
Explanation:
To find its speed, we first find the acceleration of the center of mass of a rolling object is given by
a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²
a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)
= 9.8 m/s²sin4(1 + 2/3)
= 9.8 m/s²sin4 × (5/3)
= 1.14 m/s²
To find its speed v after rolling for 60 m, we use
v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m
v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s
The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:




Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
<u>For </u>
<u>:</u>
The only force acting On the
box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.


= 915kg
<u>For </u>
<u>:</u>
There are two forces acting on
: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:


= 605kg
<u>For </u>
<u>:</u>


= 864.5kg