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Rudik [331]
3 years ago
12

The high-speed police chase ends at an intersection as a 2,150-kg Ford Explorer (driven by Robin) traveling south at 35 m/s coll

ides with a 18,250-kg garbage truck moving east at 1.55 m/s. The Explorer and the garbage truck entangle together in the middle of the intersection and move as a single object. Determine the post-collision speed two entangled vehicles.
Physics
1 answer:
defon3 years ago
4 0

Answer:

The post-collision speed of the two entangled vehicles is 5.08m/s

Explanation:

According to the law of conservation of momentum which states that the sum of momentum of a body before collision is equal to the sum of the collision of the bodies after collision. The bodies moves with the same velovity after collision.

Mathematically, mu + MU = (m+M)v

m and M are the masses of the bodies

u and U are their respective velocities

v is their common velocity

Momentum = mass*velocity

BEFORE COLLISION;

Momentum of Ford Explorer = 2150*35

= 75,250kgm/s

Momentum of the garbage truck = 18, 250 * 1.55

= 28,287.5kgm/s

AFTER COLLISION;

Note that the body will move with the same velocity after collision.

Momentum of the bodies after entangling = (2,150+18,250)v

= 20,400v kgm/s (v is their common velocity)

According to thr law;

75,250+28,287.5 = 20,400v

103,537.5 = 20,400v

v = 103,537.5/20,400

v = 5.08m/s

The post-collision speed of the two entangled vehicles is 5.08m/s

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Mechanical advantage allows you to apply a force over a short distance to increase the distance and object moves.

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A passenger on your boat falls overboard. what should you do first?
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The very first thing that you should do when a passenger on your boat falls overboard is to throw a PFD or also known as a Personal Flotation Device. This would include anything that can help the passenger to float. But this step would differ, only if the passenger is not wearing a lifevest. 
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3 years ago
A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image
lilavasa [31]

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.  T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0
3 years ago
A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

5 0
2 years ago
Read 2 more answers
An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

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Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

7 0
2 years ago
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