(a) Is the velocity of car A greater than, less than, or the same as thevelocity of car B?
(b) Is the initial position of car A greater than, less than, or equal to the
initial position of car B?
(c) In the time period from t = 0 tot = 1 s, is car A ahead of car B,
behind car B, or at the same position as car B?

Answer:

See answer below

Explanation:

Hi there,

To get started, recall the Center of Mass formula for two masses:

 where m is mass and x is displacement from the center of the shape.

Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.

We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:

Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.

Study well and persevere.

thanks,

Final answer:

To balance the rod with the attached objects, place a support at a distance of L/3 from the left end of the rod.

Explanation:

To balance the rod with the attached objects, you need to place a support at a distance of L/3 from the left end of the rod. This is because the center of gravity of the system should be directly above the support.

The center of gravity is given by the equation xcg = (m*0 + m*L/2 + 2m*L)/ (m + m + 2m). Solving this equation, we get xcg = 2L/3. Therefore, the support should be placed at a distance of L/3 from the left end of the rod.

Learn more about Balancing a rod with objects here:

brainly.com/question/16053283

#SPJ3

Answer:

96.05 N

Explanation:

From Vector,

The two forces acting along the x and y axis are perpendicular,

Fr = √(60²+75²) .............. Equation 1

Where Fr is the result of the two forces

Fr = √(3600+5625)

Fr = √(9225)

Fr = 96.05 N.

Note: Since the object moves with a constant velocity when it is acted upon by the three forces, The acceleration is zero and as such the resultant of the forces is equal to zero.

Therefore,

Ft = Fr+F3................... Equation 2

Where Ft = Total resistance of the three forces, F3 = magnitude of the third force.

make F3 the subject of the equation,

F3 = Ft-Fr

Given: Ft = 0 N, Fr = 96.05 N.

Substitute into equation 2

F3 = 0-96.05

F3 = -96.05 N.

Answer:

m¹=30.3kg

m²=40.17kg

R=0.5m

G=6.67*10¹¹

F=Gm¹m²/R²

=160.68

Answer:

1. 1 s
2. 19.6 m
3. 2 s
4. 0.8 m/s^2
5. 28 m/s
6. 79 m/s
7. 0.37 s
8. 26 m/s
9. 242 m/s
10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

Additional comment

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

The man can climb , before  the ladders starts to slip.  

A - point at the top of the ladder  

B - point at the bottom of the ladder  

C - point where the man is positioned in the ladder  

L- the length of the ladder  

α - angle between ladder and ground  

x - distance between C and B  

The forces act on the ladder,  

Horizontal reaction force (T) of the wall against the ladder  

Vertical (upward) reaction force (R) of ground against the ladder.  

Frictionalhorizontal ( to the left ) force (F)  

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N  

in static conditions,  

∑Fx = T - F = 0                   Since,  T = F  

∑Fy = mg - R = 0                Since,  735 - R = 0, R = 735  

∑ Torques(b) = 0  

In point B the torque produced by forces R and F is Zero  

Then:  

∑Torques(b) = 0        

And the arm lever for each force,  

mg = 735  

Since, ∑Torques(b) = 0    

      Since,T = F  

F < μR the ladder will starts slipping over the ground  

μ(s) = 0.25    

Therefore, the man can climb , before  the ladders starts to slip. \

To know more about  Torque,

brainly.com/question/6855614  

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call  

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right)  reaction (T) of the wall against the     ladder

Point B : Vertical (upwards) reaction (R)  of ground against the ladder

               frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s²       mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0    ⇒   T = F

∑Fy = mg - R = 0       ⇒   735 - R = 0     ⇒  R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0      

And the arm lever for each force is:

mg = 735

d₁ for mg     and d₂  for T

cos α = d₁/x     then    d₁ = x*cosα

sin α  = d₂ / L   then    d₂ = L*sinα

Then:

∑Torques(b) = 0     ⇒   735*x*cosα  - T*L*sinα = 0

735*x*cosα =  T*L*sinα

T = F then       735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα         cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L     or   F = (735)*x/L*tanα

When F < μ* R  the ladder will stars slippering over the ground

μ(s) = 0,25           0,25*R = 735*x/L*tanα

x   = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then  x (max) = 0,25*L*tanα

Answer:

B

Explanation: