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Gekata [30.6K]
3 years ago
6

Which expression should appear in the blue box to compute percent error? Accepted value – Experimental value Accepted value × Ex

perimental value Experimental value + Accepted value Experimental value – Accepted value
Physics
1 answer:
Alexxx [7]3 years ago
5 0
The last choice ... Experimental value -- <span>Accepted value ... is the place
to start, but it doesn't quite get you there.

To find the FRACTIONAL error, you'd have to divide that difference
by the Accepted Value.


And then . . .

To find the PERCENTAGE error, you'd have to multiply
the fractional error by 100 .
</span>
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Explain why cars are designed so the air pushes them towards the ground when they travel at high speed.​
Studentka2010 [4]

Answer:

gravity, refer to le mans incident

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2 years ago
The length of an aluminum wire is quadrupled and the radius is doubled. By which factor does the resistance change?
erik [133]
The resistance of a conductive wire is given by:
R= \frac{\rho L}{A}
where
\rho is the material resistivity
L is the wire length
A is the cross-sectional area of the wire

The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write 
L'=4 L

Similarly, the radius of the wire is doubled (r'=2r), so the new area is
A'= \pi (r')^2 = \pi (2r)^2 = 4 \pi r^2 = 4A

And if we substitute into the equation, we find that the new resistance of the wire is
R'= \frac{\rho L'}{A'}= \frac{\rho (4L)}{4 A'}  =  \frac{\rho L}{A}=R
Therefore, R=R': this means that the resistance of the wire did not change.
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Korvikt [17]

Explanation:

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3 0
3 years ago
Read 2 more answers
What does a halogen atom give off when it gains an electron?
spin [16.1K]
Probably gas because it will reach Nobel gas state
6 0
3 years ago
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Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg
n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N

Thus, the net force over the body is:

F=(913.14N)\hat{i}+(274.87N)\hat{j}

Next, you calculate the magnitude of the force:

F=\sqrt{(913.14N)+(274.87N)^2}=953.61N

and the direction is:

\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°

7 0
3 years ago
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