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skad [1K]
3 years ago
8

1) Roland has built a circuit, and is using a device called an ammeter to measure how quickly electrical current is flowing thro

ugh the circuit. He calculates that the current should be 0.1800.180 amps, but he measures the current as 0.1730.173 amps. What is Roland's percent error? 2) Two objects have masses of 1.1551.155 kg and 0.220.22 kg, respectively. Using the correct number of significant figures, what is the sum of these masses? 3) Kwame is measuring the dimensions of a box he will be using in a coming experiment. He measures the dimensions of the box as 0.8850.885 meters wide 0.200.20 meters deep, and 0.750.75 meters tall. Using the correct number of significant figures, what is the volume of this box? please try to answer as many questions as you can, thank you.
Physics
1 answer:
zaharov [31]3 years ago
6 0

Answer:

1)  3.9 %

2) 1,38 kg

3) 0.13 cubic meters

Explanation:

Part 1)

Calculated current = 0.180 amps

Measured current: 0.173 amps

percent error :   \frac{0.180 - 0.173}{0.180} \,100=3.9 %

Part 2)

mass1 = 1,155 kg

mass2 = 0.22  kg

Addition; 1.155 + 0.22 = 1 375 kg  which should be rounded to just two decimals according to the addition rules for uncertainties:

Answer : 1,38 kg

Part 3)

Volume = 0.885 x 0.20 x 0.75 cubic meters = 0.13275 cubic meters

Notice that the smallest number of significant figures among the set of dimension given is two. Therefore, our answer should be rounded to two significant figures. That is:

Volume = 0.13 cubic meters

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anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

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F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and
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Answer:

The maximum shear stress in shaft AB, T_{ABmax} is 15 MPa

The maximum shear stress in shaft CD,  T_{CDmax} is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = \pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}

Therefore, we have

J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}

Where:

D_{AB} = Diameter of shaft AB = 30 mm = 0.03 m

r_{AB} = Radius of shaft AB = 15 mm = 0.015 m

∴ J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2} = 7.95 × 10⁻⁸ m⁴

and

J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}

Where:

D_{CD} = Diameter of shaft CD = 36 mm = 0.036 m

r_{CD} = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2} = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= 1.58 \times \frac{2\pi }{360} rad = 2.76 × 10⁻² rad.

That is \phi_r = 2.76 × 10⁻² rad.

However  \phi_r =  \phi_{C/D} -  \phi_{B/A}  

Where:

\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G} and

\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}

T_{AB} and T_{CD}= Torque on shaft AB and CD respectively

T_{AB}  = Required

T_{CD}= 500 N·m

L_{AB} and L_{CD} = Length of shafts AB an CD respectively

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G = Shear modulus of the material = 77.2 GPa

Therefore;

\phi_r =  \phi_{C/D} -  \phi_{B/A}  =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

2.76 × 10⁻² rad =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

=\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}

Therefore;

T_{AB} =  79.54 N.m

Where T = T_{AB} + T_{CD} =

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τ_{max} = \frac{T\times R}{J}

\tau_{ABmax} = \frac{T_{AB}\times R_{AB}}{J_{AB}} =  \frac{79.54\times 0.015}{7.95\times 10^{-8}} = 15 MPa

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A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
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