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3 years ago

When determining the number of significant digits in a measurement,

Physics

2 answers:

Black_prince [1.1K]3 years ago

6 0

B) All nonzero digits are significant.

myrzilka [38]3 years ago

3 0

Answer:

all nonzero digits are significant.

Explanation:

Significant digits are total number of digits that are used to define the final results.

So here we have to define as per rules

1) All non zero numbers are significant

2) All zeros between two non zero numbers are significant

3) All trailing zeros in a number without decimal are not significant

4) All zeros in a number which is greater than 1 with decimal are significant

So as per above rules correct answer will be

all nonzero digits are significant.

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3 years ago

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A disk has a radius of 30 cm and a mass of 0.3 kg and is turning at 3.0 rev/s. A trickle of sand falls onto the disk at a distan

Pani-rosa [81]

Answer:

The mass of the sand that will fall on the disk to decrease the is 0.3375 kg

Explanation:

Moment before = Moment after

$I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i - I \omega_f \\\\m = \frac{ I \omega_i - I \omega_f}{r^2 \omega_f }$

where;

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substitute this in the above equation;

$m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg$

Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg

7 0

3 years ago

2 kg

pochemuha

Answer:

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3 years ago

A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin

anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

$\Delta L = L_0 \alpha \Delta T$

Where,

$L_0 =$ Initial Length

$\alpha =$ Thermal expansion coefficient

$\Delta T =$ Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

$0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C$

$103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C$

The coefficient of thermal expansion given is

$\alpha = 1.1*10^{-5}/\°C$

The initial length would be,

$L_0 = 565m$

Replacing we have to,

$\Delta L = L_0 \alpha \Delta T$

$\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))$

$\Delta L = 0.355m$

This means that the building will be 35.5cm taller

3 0

3 years ago