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Zinaida [17]
2 years ago
14

2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de

termine the distance from the edge of the well to the water's surface.​
Physics
1 answer:
kow [346]2 years ago
3 0

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

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The solution for this problem is:
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= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer
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2 years ago
Mrs. Gonzalez is about to give birth and Mr. Gonzalez is rushing her to the hospital at a speed of 30.0 m/s. Witnessing the spee
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Answer: The frequency = 1714.3Hz

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2 years ago
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
SVETLANKA909090 [29]

Answer:

The  value is  v =  -0.04 \  m/s

Explanation:

From the question we are told that

   The  mass  of the block is  m  =  2.0 \ kg

   The  force constant  of the spring is  k  =  590 \ N/m

   The amplitude  is  A =  + 0.080

   The  time consider is  t =  0.10 \  s

Generally the angular velocity of this  block is mathematically represented as

      w =  \sqrt{\frac{k}{m} }

=>   w =  \sqrt{\frac{590}{2} }

=>   w = 17.18 \  rad/s

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         v  =  -A w sin (w* t )

=>       v  = -0.080 * 17.18 sin (17.18* 0.10 )

=>       v =  -0.04 \  m/s

7 0
2 years ago
A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far d
AlekseyPX

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

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1 year ago
Identify the method of thermal energy transfer at work in hot air balloons. Explain how thermal energy is transferred in this sc
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Thermal energy in the form of fire is generated by the combustion of fuel. Due to the tendency of hot air to rise upward, the heat generated rises to fill the space of the balloon. One this space is full of trapped hot air, the heat's tendency to rise causes the hot air balloon to be lifted into the air. 
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