The fluid pressure 10 ft underwater is ____ the fluid pressure 5 ft underwater

What is the velocity of a wave that has a frequency of 400Hz and a wavelength of 0.5 meters

iren2701 [21]

Answer: 12

Explanation:

Let’s take for instance the case of a wave with a frequency of 400 Hz going through a material at a speed of .5 m/s. The wavelength result is 12 m. Wave velocity (m/s) = Frequency (Hz) x Wavelength (m)

4 0

2 years ago

Read 2 more answers

A marble is placed at Location W. When the marble is released, it rolls down the track.

Norma-Jean [14]

Answer:

Z

Explanation:

6 0

2 years ago

Read 2 more answers

what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N

MArishka [77]

Explanation:

6gtvctyvyvyvyvyfcfufy

8 0

2 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

6 0

2 years ago

Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-3.4 μC). Charge q2 is located at

ivanzaharov [21]

I believe the answer would be zero because the q1 and q2 are equal on opposite sides and it

hope this helps

3 0

2 years ago