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zavuch27 [327]
3 years ago
5

A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?

Physics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

option A

Explanation:

given,                                              

height of the drop of stone = 9.44 m

speed of the stone = ?                          

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.            

Potential energy = Kinetic energy    

m g h = \dfrac{1}{2} m v^2  

     v = \sqrt{2gh}                  

     v = \sqrt{2\times 9.8 \times 9.44}

     v = \sqrt{185.024}              

            v = 13.60 m/s                      

Hence, the correct answer is option A

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Below is an "oracle" function. An oracle function is a function presented interactively. When you type in a t value, and press t
Bingel [31]

Answer:

The rate of change of height with respect to time is -10.64 feet/sec

Explanation:

Given that,

There are three lines, so you can calculate three different values of the function at one time.

The function f(t) represents the height in feet of a ball thrown into the air, t seconds after it has been thrown.

Given table is,

Time t = 0, 1, 1,02

Function is,F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015

We need to calculate the initial height of ball

Using equation of motion

h=h_{0}+ut-\dfrac{1}{2}gt^2

Where, h₀ = initial height

u = initial velocity

t = time

g = acceleration due to gravity

At t = 0,

Put the value into the formula

-3.053113177191196\times10^{-18}=h_{0}+0-0

h_{0}=-3.053113177191196\times10^{-18}

We need to calculate the height of ball at t = 1

Using equation of motion

h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2

Put the value in the equation

6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32

6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}

u_{0}=22\ feet/s

Velocity is the rate of change of height with respect to time

So, velocity at 1.02 sec is given

We need to calculate the height

Using equation of motion

h=ut-\dfrac{1}{2}gt^2

On differentiating w.r.to t

h'(t)=u-\dfrac{1}{2}g(2t)

Put the value into the formula

h'(t)=22-\times32\times(1.02)

h'(t)=-10.64\ feet/sec

Hence, The rate of change of height with respect to time is -10.64 feet/sec.

4 0
3 years ago
A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
3 years ago
Question 5 of 32
harkovskaia [24]
The answer is D
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a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration
slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0
3 years ago
Mass of block A above is 25-kg. Calculate the force of gravity on block A.
stiv31 [10]

Answer:w+mg

Explanation:

4 0
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