A net torque of magnitude 600 Nm is ex-

Determine if the data are qualitative, quantitative, or neither. Zinc is a silver-gray metal. Chlorine has a density of 3.2 g/L.

AysviL [449]

Qualitative data gives the information of quality which can not be measured in numbers. For example: Color of eyes, softness of skin.

Quantitative data is information of quantity that can be represented in numbers. For example length and mass of any object.

Zinc is a silver-gray metal is a qualitative data, here silver gray color is quality of zinc metal which can not be measured in numbers.

Chlorine has a density of 3.2 g/L is a quantitative data. The value of density can be compared with other elements by comparing the numbers.

Gallium is not found in nature is neither qualitative nor quantitative.

Nitrogen has a melting point of –210.00 °C is a quantitative data because this is expressed in numbers.

Aluminum is a solid is a qualitative data because it tells about the state of element which can not be measured in numbers.

5 0

3 years ago

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The radius of curvature of a spherical mirror is 20cm.What is its focal length?

hichkok12 [17]

Answer:

$\boxed{\sf Focal \ length = 10 \ cm}$

Given:

Radius of curvature (R) of a spherical mirror = 20

To Find:

Focal length (f)

Explanation:

Formula:

$\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}$

Substituting value of R in the equation:

$\sf \implies f = \frac{20}{2}$

$\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}$

$\sf \implies f = 10 \: cm$

5 0

3 years ago

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What is a bond between a positive and a negative ion charge?

klasskru [66]

Answer:

Ionic Bond

Explanation:

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2 years ago

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A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t

kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

$\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W$

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

$emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV$

hope this helps!!

5 0

3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

3 years ago