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Andrej [43]
3 years ago
10

The melting point of potassium thiocyanate determined by a student in the laboratory turned out to be 174.5 oC. The accepted val

ue of this melting point is 173.2 oC. What is the percent error in this reading?
Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

0.75%

Explanation:

Measured value of melting point of potassium thiocyanate = 174.5 °C

Actual value of melting point of potassium thiocyanate = 173.2 °C

<em>Error in the reading = |Experimental value - Theoretical value|</em>

<em>= |174.5 - 173.2|</em>

<em>= |1.3|</em>

<em>Percentage error = (Error / Theoretical value) × 100</em>

<em>= (1.3 / 173.2)×100</em>

<em>= 0.75 %</em>

∴ Percentage error in the reading is 0.75%

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Answer:

An object gets charged when it's atoms lose or gain an electron to become an ion. For example: ... This means that the fur loses it's electrons to the plastic rod and both objects are now charged. The fur is positively charged because it lost electrons and the rod is negatively charged because it gained electrons

Explanation:

comment how it helps

3 0
3 years ago
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An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the
ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.    

The magnetic field in north direction, B_y=19911.5\ nT

The magnetic field in west direction, B_x=-3257.1\ nT

The magnetic field in vertical direction, B_z=48381.8 \ nT

Magnetic field, B=(-3257.1i+19911.5j+48381.8k)\ nT

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

E_k=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E_k}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}

v=5.92\times 10^7 i\ m/s (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

F=q(v\times B)

F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})

Since, i\times j=k\ \\j\times k=i\\k\times i=j

And, i\times i=j\times j=k\times k=0

F=1.6\times 10^{-19}\times [1178 k-2864.20j]

|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}

F=4.95\times 10^{-16}\ N

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a=\dfrac{F}{m}

a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}

a=5.43\times 10^{14}\ m/s^2

Hence, this is the required solution.

4 0
3 years ago
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates
marta [7]
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asambeis [7]
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So, the correct answer would be 30 AU. Hoped this helped!
7 0
3 years ago
PLS HELP QUICK TYSM!!!!!!!!!!
maksim [4K]

C. 4 N is the answer if not then i be dumb!!!

8 0
4 years ago
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