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omeli [17]
3 years ago
9

A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a

person stands at the tree and pulls ten feet of the rope (after all slack is taken up) how far will the vehicle move?

Physics
1 answer:
Angelina_Jolie [31]3 years ago
3 0

Answer:

The vehicle displacement is 9.90 feet.

Explanation:

Given that,

Height of tree = 4.5 feet

Distance = 33 feet

According to figure,

We need to calculate the value of l

Using Pythagorean theorem

l=\sqrt{(33)^2+(4.5)^2}

We need to calculate the vehicle displacement

Using horizontal component

Vehicle displacement =horizontal component of pulled rope

Vehicle\ displacement= d\cos\theta

Where, \theta is angle between rope and ground

d = pulled length of rope

Vehicle\ displacement=10\times\dfrac{33}{\sqrt{(33)^2+(4.5)^2}}

Vehicle\ displacement=9.90\ feet

Hence, The vehicle displacement is 9.90 feet.

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A 'displacement' always consists of a magnitude and a direction.  The two cars you just described have displacements with the same magnitude ... 5 km.  But if they didn't both drive in the same direction, then their displacements are different.

Remember:

-- 10 m/s² up and 10 m/s² down are different accelerations

-- 30 mph East and 30 mph West are the same speed but different velocity.

-- 5 km North and 5 km South are the same distance but different displacement.

7 0
3 years ago
A. State whether the acceleration is positive, negative or zero for each section in the speed versus
bazaltina [42]

Answer:

a=positive

b=0

c=positive

d=negative

Explanation:

a=acceleration depends on the speed and time. if the speed and time are increasing at the same rate, the acceleration value will be positive as the vehicle is speeding up.

b=the speed and time are not increasing, therefore the vehicle is either stationary or travelling at a steady pace.

c=same explanation as a

d=the speed and time are not increasing at the same rate as the speed is decreasing. this means that the car is slowing down

5 0
2 years ago
A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
emmainna [20.7K]

Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  H = 27 .67 \ m

The  pumpkin's vertical velocity when it hits the ground is  v_v__{f}} = 23.298 \  m/s

Explanation:

From the question we are told that

   The  horizontal speed is  v_h  =  4 m/s

    The horizontal distance traveled is  d =  9.5 \ m

The horizontal distance traveled is mathematically represented as

           S =  v_h * t

Where t is the time taken

substituting values

          9.5 =  4 * t

   =>     t =  \frac{9.5}{4}

            t = 2.38 \ sec

Now the vertical displacement is mathematically represented as

        H  =  v_v t  +  \frac{1}{2} a_v t^2

now the vertical velocity before the throw is  zero

    So

          H =  0 +  \frac{1}{2} (9.8) * (2.375)^2

          H = 27 .67 \ m

Now the final vertical velocity  is mathematically represented as

          v_v__{f}} =  v_v + at

  substituting values

             v_v__{f}} =  0 + (9.8)* (2.375)

            v_v__{f}} = 23.298 \  m/s

7 0
3 years ago
Two cars are traveling on a desert road. After 5.0 seconds they are side by side at the next telephone pole.The distance between
stira [4]
The question looks incomplete, but according to the information given above seem like they have <span>identical journeys.
</span>a. the displacement of car A - <span> 65.5 m 
</span>b. the displacement of car B  - <span>65.5 m
c. average velocir</span>y of A  65.5 / 5 = 13.1 m/s
d. the average velocity of car B  has the same. 
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3 years ago
Which statement describes a question that can be answered by a scientific
goblinko [34]
It’s d! hope I can help
4 0
3 years ago
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