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antiseptic1488 [7]
3 years ago
5

Mercury is a metallic liquid element. It has a density of 11.3 g/cm3 . If you placed the metals listed above in mercury, which o

f the choices below describes what would happen? A - All would float. B - Gold would float, the others would sink. C - All would float except gold. D - All would sink
Physics
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

The answer is below

Explanation:

The question is not complete since the liquid density is not given.

Archimedes principle states that a body at rest in a fluid is acted upon by an upward force known as the buoyant force. The buoyant force is equal to the weight of the fluid displaced.

An object floats when it is placed in a liquid only if the density of the object is less than the density of the liquid. Therefore those metals with density less than that of mercury would float while those with density greater than mercury would sink.

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A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t
4vir4ik [10]

The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force  is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is  by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

rms voltage = 78 K v

power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

Learn more about rms current here:-brainly.com/question/20913680

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7 0
2 years ago
A cylindrical container closed of both end has a radius of 7cm and height of 6cm A.)find the total surface area of the container
Natalka [10]

Given:

A cylindrical container closed of both end has a radius of 7cm and height of 6cm.

Explanation:

A.) Find the total surface area of the container.

  • A = 2πrh + 2πr²
  • A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
  • A = 263.76 + 307.72
  • A = 571.48

B.) Find the volume of the container.

  • V = πr²h
  • V = (3.14)(7×7)(6)
  • V = 923.16

Not sure huhuness.

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8 0
3 years ago
Read 2 more answers
Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.
Firlakuza [10]

Answer:

s = 414.7 m\\

Explanation:

The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:

s = rθ

where,

s = distance traveled on road = ?

r  radius of tires = diameter/2 = 2.2 m/2 = 1.1 m

θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad

Therefore,

s = (1.1 m)(377 rad)\\s = 414.7 m

8 0
3 years ago
An object traveling at 1.5 rad
Veronika [31]

The object's final velocity, given the data is 10.5 rad/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

The following data were obtained from the question

  • Initial velocity (u) = 1.5 rad/s
  • Acceleration (a) = 0.75 rad/s²
  • Time (t) = 12 s
  • Final velocity (v) = ?

The final velocity can be obtained as follow:

a = (v – u) / t

0.75 = (v – 1.5) / 12

Cross multiply

v – 1.5 = 0.75 × 12

v – 1.5 = 9

Collect like terms

v = 9 + 1.5

v = 10.5 rad/s

Thus, the final velocity of the object is 10.5 rad/s

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brainly.com/question/491732

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6 0
2 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
2 years ago
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