All categories

3 years ago

When you observe a physical property the substance

2 answers:

5 0

<span>The observation or measurement of physical properties of matter does not change its composition or its chemical nature. Other examples of physical properties include the infrared spectrum, attraction or repulsion to magnets, viscosity and opacity.</span>

Blizzard [7]3 years ago

5 0

Of what? what is the substance

You might be interested in

A 40-w lightbulb connected to a 120-v source experiences a voltage surge that produces 132 v for a moment. by what percentage do

andrey2020 [161]

R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%

3 0

3 years ago

A ____ is one of the tiny dots of light that form a grid on your screen.

Degger [83]

Hi there

The answer is pixel

Hope this helps

8 0

3 years ago

Read 2 more answers

A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. Find the magnitude of the

Fittoniya [83]

Answer : 0.814 newton

Explanation:

force (magnetic) acting on the wire is given by

F= ? , I=2.2amp , B = 0.37 T

F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N

4 0

3 years ago

Read 2 more answers

An inductor is connected to a 26.5 Hz power supply that produces a 41.2 V rms voltage. What minimum inductance is needed to keep

alexira [117]

Answer:

The minimum inductance needed is 2.78 H

Explanation:

Given;

frequency of the AC, f = 26.5 Hz

the root mean square voltage in the circuit, $V_{rms}$ = 41.2 V

the maximum current in the circuit, I₀ = 126 mA

The root mean square current is given by;

$I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_{rms} = \frac{126*10^{-3}}{\sqrt{2} }\\\\I_{rms} =0.0891 \ A$

The inductive reactance is given by;

$X_l = \frac{V_{rms}}{I_{rms}} \\\\X_l= \frac{41.2}{0.0891}\\\\X_l = 462.4 \ ohms$

The minimum inductance needed is given by;

$X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2\pi f}\\\\L = \frac{462.4}{2\pi *26.5}\\\\L = 2.78 \ H$

Therefore, the minimum inductance needed is 2.78 H

7 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago