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SSSSS [86.1K]
2 years ago
12

At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water th

at is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25° C. Suppose that a temperature increase of more than 18° C is considered to be potentially dangerous to the environment. What is the probability that at any point in time, the temperature increase is potentially dangerous?
Physics
1 answer:
Mice21 [21]2 years ago
8 0
Answer is 30. just took it
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Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
2 years ago
A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
Nina [5.8K]

Answer: SG = 2.67

Specific gravity of the sand is 2.67

Explanation:

Specific gravity = density of material/density of water

Given;

Mass of sand m = 100g

Volume of sand = volume of water displaced

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Therefore, the specific gravity of sand is

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

Specific gravity of the sand is 2.67

3 0
2 years ago
Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.
Katena32 [7]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The power created  is  P_{avg} =  F  *  v_{avg}

Explanation:

From the question we are told that

    The that the average power is  mathematically represented as

            P_{avg} =  \frac{W }{\Delta  t  }

Where W is  is the Workdone which is  mathematically represented as

         W =  F  *  s

      Where F is  the applies force and  s  is the displacement  due to the force  

        So  

                P_{avg} =  \frac{F *s  }{\Delta  t  }

Now this  displacement can be represented mathematically as  

            s =  v_{avg} *  \Delta  t

Where v_{avg } is the average  velocity and \Delta  t is the time  taken  

So  

            P_{avg} =  \frac{F *v_{avg} *  \Delta  t    }{\Delta  t  }

=>         P_{avg} =  F  *  v_{avg}

3 0
2 years ago
A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus
dexar [7]

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

6 0
2 years ago
Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown
olasank [31]

Answer:

a)  v = √ g x , b)  W = 2 m g d , c)    a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

            T-W₁ = m a

            W₃ - T = M a

            w₃ - w₁ = (m + M) a

            a = (3m - m) / (m + 3m) g

            a = 2/4 g

            a = ½ g

the speed of the blocks is

          v² = v₀² + 2 ½ g x

          v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

           W₁ = -W d = - mg d

3m heavy block

             

            W₂ = W d = 3m g d

the total work is

             W = W₁ + W₂

             W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

8 0
2 years ago
Read 2 more answers
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