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Tcecarenko [31]
3 years ago
10

Which of the following describes density only in terms of metric units?

Physics
1 answer:
Alborosie3 years ago
7 0
A) Kilograms per cubic meter. Every other option either contains pounds or feet, which are both units of measurement from the standard system, not the metric system.
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A 100-m-long wire carrying a current of 4.0 A will be accompanied by a magnetic field of what strength at a distance of 0.050 m
solong [7]

Answer:

1.6 x 10^-5 T

Explanation:

i = 4 A

r = 0.05 m

The magnetic field due to long wire at a distance r is given by

B = \frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B = 10^-7 x 2 x 4 / 0.05

B = 1.6 x 10^-5 T

3 0
3 years ago
A figure shows a vertically moving block on the end of a cord. The graph next to the figure gives the vertical velocity componen
Sergio [31]
It should be 2/9/10 in functional form
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3 years ago
A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor
natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = 120\Omega

Fixed resistance, R = 120\Omega

Gauge Factor, G.F = 2.01

Supply Voltage, V_{s} = 3\ V

Strain, \epsilon = 1000\times 10^{-6}\ strain

Now,

To calculate the output voltage, V_{o}:

WE know that strain is given by:

\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}

Thus

V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}

Now, substituting the suitable values in the above eqn:

V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}

V_{o} = 1.507\ mV

6 0
3 years ago
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
Greeley [361]

Answer:

I = 215.76 A  

Explanation:

The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.  

Fm = u_o*I^2*L/2*π*d

where I is the current in each rod, d = 0.0082 m is the distance 27rId  

between each, L = 0.85 m is the length of each rod.

Fm = 4π*10^-7*I^2*1.1/2*π*0.0083

The mass of each rod is m = 0.0276 kg  

F_m = mg

4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8

I = 215.76 A  

note:

mathematical calculation maybe wrong or having little bit error but the method is perfectly fine

5 0
3 years ago
g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
2 years ago
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