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4 years ago

10

If your volume measurement is only good to two significant figures, will your results be more precise if you use mass measuremen

ts that have six significant figures. Is it true?

1 answer:

Rina8888 [55]4 years ago

4 0

Answer:

Measurement with six significant figure has a possibility of much greater accuracy than the measurement of only two significant figure.

Explanation:

When we measure an object in only two significant figures then we eliminate the possibility of getting close to higher accuracy as when the measurement is taken into six significant figures.

When we measure in six significant figures then we take the measurement into more finer fragments, we are able to measure very smaller variation as well. In other words we are taking the least count of the measurement to a more finer level than when measuring upto two significant figures.

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A calculator has a resistance of 22 Ω. What is the power rating for this calculator when connected to a 1.5 V battery?

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Answer:

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Explanation:

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3 years ago

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3 years ago

A ball whose mass is 0.4 kg hits the floor with a speed of 8 m/s and rebounds upward with a speed of 6 m/s. Collapse question pa

Colt1911 [192]

Answer:

The the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

Explanation:

Given:

Mass of ball $m = 0.4$ kg

Initial speed $v_{i} = -8 \frac{m}{s}$

Rebound speed $v_{f} = 6 \frac{m}{s}$

Contact time interval $\Delta t = 0.5 \times 10^{-3}$ sec

For finding the average magnitude of the force on the ball by the floor is given by,

$F_{avg} = \frac{\Delta P}{\Delta t}$

Here $\Delta P = m (v_{f}- v_{i} )$

$F_{avg} = \frac{m (v_{f} -v_{i} )}{\Delta t}$

$F_{avg} = \frac{0.4 \times (6 -( -8 ) )}{0.5 \times 10^{-3} }$

$F_{avg} = 11.2 \times 10^{3}$ N

Therefore, the the average magnitude of the force exerted on the ball by the floor $11.2 \times 10^{3}$ N

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3 years ago

Read 2 more answers

Newton's laws of motion work well for ordinary situations on earth. However, these laws of motion do not work for all cases. Und

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4 years ago

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Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold

arsen [322]

Answer:

Resistance of gold wire, $R=1977 \times 10^3 ohm$

Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$ .........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

$A=2.53\times 10^{-11}m^2$

we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$ ................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

7 0

3 years ago