Which equations represent the relationship between wavelength and frequency for a sound wave? Check all that apply.

An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.

Minchanka [31]

Answer:

Fₓ = 0, $F_{y}$ = 0 and $F_{z}$ = - 3.115 10⁻¹⁵ N

Explanation:

The magnetic force given by the expression

F = q v xB

the bold are vectors, the easiest analytical way to determine this force in solving the determinant

$F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}$

F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

F =i^0 + j^0 - k^ 3.115 10⁻¹⁵ N

Fₓ = 0

$F_{y}$ = 0

$F_{z}$ = - 3.115 10⁻¹⁵ N

6 0

3 years ago

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

(c) the maximum acceleration of the object

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

What does it mean if an object is in motion?

Alina [70]

Explanation:

basically it's moving I think

6 0

3 years ago

710 mm= ______________cm

love history [14]

Answer:

71 cm

Explanation:

Every 100 mm is equal to 10 cm. Hope this helps!

4 0

3 years ago

Read 2 more answers

Reactance Frequency Dependence: Sketch a graph of the frequency dependence of a resistor, capacitor, and inductor. RLC Circuit R

jolli1 [7]

Answer:

$f=\frac{1}{2\pi \sqrt{LC}}$

Explanation:

We know that impedance of a RLC circuit is given by $Z=R+J(X_L-X_C)$

So $Z=\sqrt{R^2+(X_L-X_C)^2}$ here R is resistance $X_L$ is inductive reactance and $X_C$ is capacitive reactance

To minimize the impedance $X_L-X_C$ should be zero we know that $X_L=\omega L\ and \ X_C=\frac{1}{\omega C}$

So $\omega L-\frac{1}{\omega C}=0$

$\omega ^2=\frac{1}{LC}$

$\omega =\sqrt{\frac{1}{LC}}$

We know that $\omega =2\pi f$

So $\omega =2\pi f=\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

Where f is resonance frequency

8 0

3 years ago