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3 years ago

Simlarity between longitudanal and transverse waves

1 answer:

8 0

For transverse waves, the waves move in perpendicular direction to the source of vibration. For longitudinal waves, the waves move in parallel direction to the source of vibration . They are similar in the sense that energy is transferred in the form of waves.

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11th grade physics problem...

Rus_ich [418]

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5 0

3 years ago

If Vx = 7.00 units and Vy = -7.60 units, determine the magnitude of V⃗ .

Juliette [100K]

|V| = 10.33 units and the direction θ = -47.35° or 312.65°.

Given the x and y components of a vector, we can calculate the magnitude and direction from these components.

Applying the Pythagorean theorem we have that the magnitude of the vector is:

|V| = $\sqrt{Vx^{2}+Vy^{2} }$

|V| = $\sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units$

The expression for the direction of a vector comes from the definition of the tangent of an angle:

tan θ = $\frac{Vy}{Vx}$ ------> θ = arc tan $\frac{Vy}{Vx}$

θ = arc tan $\frac{-7.60units}{7.00units}$

θ = -47.35° or 312.65°

6 0

3 years ago

Which statement is NOT true about "p" orbitals? Which statement is NOT true about "p" orbitals? A 3p orbital has a higher energy

ki77a [65]

Answer:

Option E is correction. None of the above.

Explanation:

( 1 ) A 3p orbital has more energy than 2p orbital and this is the reason it is away from the nucleus as compare to 2p orbital. Energy of the shells increases as their distance increases from the nucleus.

(2) p subshells are made up of three dumbbell-shaped orbitals

(3) There are three atomic orbitals in a p subshell. They are px, py, and pz.

3 0

4 years ago

A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

Tema [17]

Answer:

α = 1.114 × 10⁻³ (°C)⁻¹

Explanation:

Given that:

Length of rod (L) = 1.5 m,

Diameter (d) = 0.55 cm,

Area (A) = $\pi r^2$

Radius (r) = d / 2 = 0.275 cm,

Voltage across the rod (V) = 15.0 V.

At initial temperature (T₀) = 20°C, the current (I₀) = 18.8 A while at a temperature (T) = 92⁰C, the current (I) = 17.4 A

a) The resistance of the rod (R) is given as:

$R=\frac{Voltage(V)}{I_0} \\R=\frac{15}{18.8}=0.798\Omega$

Therefore the resistivity and for the material of the rod at 20 °C (ρ) is:

$\rho=\frac{RA}{L}=\frac{0.798*\pi *0.275^2}{1.5}=0.126\Omega m$

b) The temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation:

$R_T=R_0[1-\alpha (T-T_0)]\\but,R_T=\frac{V}{I}=\frac{15}{17.4}=0.862\\$

Rearranging to make α the subject of formula:

$\frac{R_T}{R_0} =1+\alpha (T-T_0)\\\alpha (T-T_0)=\frac{R_T}{R_0}-1\\\alpha =\frac{\frac{R_T}{R_0}-1}{(T-T_0)} \\Substituting:\\\alpha =\frac{\frac{0.862}{0.798}-1 }{92-20} \\\alpha =\frac{0.0802}{72} =1.114*10^-3(^0C)^{-1$

8 0

3 years ago

Charge is uniformly distributed throughout a spherical insulating volume of radius R=4.00 cm . The charge per unit volume is −6.

aleksley [76]

Answer

-8.67× 10^6 N/C

Explanation:

The Electric Field is defined as force per unit charge.

E = Q/ 4π£r2

Qv= −6.5 μCm3

Qv = Q/ V= Q/ 4/3 πr3

Hence Q = 4/3 πr3 × Qv

Hence E = 4/3 πr3 × Qv / 4π£r2= Qvr/3£

−6.5 μ × 4/ 3×8.854 ×10^-12

-6.5 × 4 × 10^6/3 = -8.67× 10^6 N/C

Note: £ = 8.854×10^-12m/F

is the permittivity of free space

Qv is the charge per unit volume

V is volume and volume

8 0

3 years ago