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vivado [14]
3 years ago
5

What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 11.0 m and the downward accel

eration of the car is 1.50 g?
Physics
1 answer:
8090 [49]3 years ago
3 0
<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration. We can use a simple equation to find centripetal acceleration. a = v^2 / r We can use this same equation to find the speed of the car. v^2 = a * r v = sqrt { a * r } v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) } v = 12.7 m/s The speed of the roller coaster is 12.7 m/s</span>
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Which of the following is responsible for the formation of sinkholes F. Eroding soil on Earths’s surface G. Ice that builds up o
monitta
Ans is <span>I. Flowing groundwater that dissolves rock , such as limestone</span>
3 0
3 years ago
The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the
Maslowich

The energy of a wave is directly proportional to the square of the amplitude of the wave.

<h3>What is the relationship between energy and amplitude?</h3>

There is direct relationship between energy of the wave and the amplitude of the wave. The energy transported by a wave is directly proportional to the square of the amplitude of the wave. This means if energy is increase the amplitude of wave becomes double and vice versa.

Energy = (amplitude)2

So we can conclude that the energy of a wave is directly proportional to the square of the amplitude of the wave.

Learn more about energy here: brainly.com/question/13881533

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3 0
1 year ago
Read 2 more answers
An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m
Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

\omega = \sqrt{\frac{k}{m}}

here we know

m = 3.5 kg

k = 270 N/m

now we have

\omega = \sqrt{\frac{270}{3.5}}

\omega = 8.78 rad/s

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

v = \omega \sqrt{A^2 - x^2}

0.55 = 8.78 \sqrt{A^2 - 0.020^2}

A = 0.066 m

Part b)

Maximum speed of SHM at its mean position is given as

v_{max} = A\omega

v_{max} = 0.066(8.78) = 0.58 m/s

4 0
3 years ago
A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
3 years ago
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
3 years ago
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