HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

If the wavelength is 10 m and the frequency is 5 Hz, what is the wave speed?

timama [110]

V = f x wavelength
V = 5 x 10
V = 50m/s

4 0

3 years ago

You run 100 meters in 50 seconds. Calculate your speed, label with m/s

NemiM [27]

Answer:

2 m/s

Explanation:

speed = distance/time

s = 100/50 = 2

4 0

3 years ago

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

work done by the gravity = ?

work done by the tension = ?

Work done by the gravity = - m g Δh

Δ h = 2 + 2 = 4 m

Work done by the gravity = $- 3 \times 9.8 \times 4$

= -117.6 J

work done by gravity is equal to -117.6 J

Work done by tension will be equal to zero.

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J

7 0

3 years ago

At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values

denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

$Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2$

but we know that at resonance $X_L=X_C$

putting $X_L=X_C$ in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

$Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}$ so from given expression it is clear that quality factor depends on R L and C

3 0

3 years ago

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

OverLord2011 [107]

Answer:

The value is $v_2 = 5.53 \ m /s$

Explanation:

From the question we are told

The pipe diameter at location 1 is $d = 8.8 \ cm = \frac{8.8 }{10} = 0.88 \ m$

The velocity at location 1 is $v_1 = 2.4 \ m /s$

The diameter at location 2 is $d_2 = 5.80 \ cm = 0.58 \ m$

Generally the area at location 1 is

$A_1 = \pi * \frac{d^2}{ 2}$

=> $A_1 = \pi * \frac{0.88^2}{ 2}$

=> $A_1 = 3.142 * \frac{0.88^2}{ 2}$

=> $A_1 = 1.2166 \ m^2$

Generally the area at location 1 is

$A_2 = \pi * \frac{d_1^2}{ 2}$

=> $A_2 = \pi * \frac{0.58^2}{ 2}$

=> $A_2 = 0.528 \ m^2$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

=> $1.2166 * 2.4 = 0.528 * v_2$

=> $v_2 = 5.53 \ m /s$

3 0

3 years ago