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andrew-mc [135]
2 years ago
9

Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An

engineer tests a cell phone and finds that the batteries supply 10,000 J of energy to make 5500 J of output energy in the form of sound and light for the screen. How efficient is the phone?
Physics
1 answer:
Svetradugi [14.3K]2 years ago
5 0


efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

                                        =55%

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You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
mamaluj [8]

The mass of an electron, me, 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them, namely numbers 9, 1, and 1

<h3><em>Further explanation</em></h3>

Significant numbers are obtained from measurement results of exact numbers and the last number estimated

This is called significant numbers

the scientific notation can be described as:

a, ... x 10ⁿ

a, ... called a significant number

10ⁿ is called a big order

Rules for significant numbers in general:

  • 1. All non-zero numbers are significant numbers
  • 2. a zero which is located between two non-zero numbers including a significant number
  • 3. all zeros are located in the final row written behind the decimal point of the include significant number
  • 4. zero decimal point is the not significant number

From the numbers known in the question amounting to 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them,(symbol a, before a big order ,called a significant number) namely numbers 9, 1, and 1

<h3><em>Learn more</em></h3>

significant figures

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significant figures in the following number: 5.67 x 106

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the number of significant figures is 0.025

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0.080 significant figures

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the fewest number of significant figures

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Keywords : significant number, the scientific notation, decimal point, a big order

6 0
1 year ago
Read 2 more answers
An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th
pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Q=mc\Delta T

Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

5 0
2 years ago
Consider an alcohol and a mercury thermometer that read exactly 0 oC at the ice point and 100 oC at the steam point. The distanc
Alexeev081 [22]

Answer:

No, both the thermometers will give the different reading.

Explanation:

Given,

  • Both thermometer has same ice point = T_i\ =\ 0^o C
  • Both thermometer has same steam point = T_s\ =\ 100^o C
  • Distance between the ice point and steam point in both the thermometer is same of 100 division,

All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion.  Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.

7 0
1 year ago
A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a
Yanka [14]

Answer:

Diameter of the cylinder will be d=2.998\times 10^4m

Explanation:

We have given young's modulus of steel E=207GPa=207\times 10^9Pa  

Change in length \Delta l=0.38mm

Length of rod l=500mm

Load F = 11100 KN

Strain is given by strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}

We know that young's modulus E=\frac{stress}{strain}

So 207\times 10^9=\frac{stress}{7.6\times 10^{-4}}

stress=1573.2\times 10^{-5}N/m^2

We know that stress =\frac{force}{artea }

So 1573.2\times 10^{-5}=\frac{11100\times 1000}{area}

area=7.055\times 10^{8}m^2

So \frac{\pi }{4}d^2=7.055\times 10^{8}

d=2.998\times 10^4m          

6 0
2 years ago
26. An ice-skater who weighs 200 N is gliding across the ice. If the force of friction is 4 N. what is the
Scrat [10]

Answer:

0.02

Explanation:

coefficient of kinetic friction = μ

force of friction = Ff

Normal Force = FN, but

FN = -W

Ff = -μFN

so μ = Ff/FN

= 4N/200N

= 0.02.

7 0
2 years ago
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