Ask question

Ask question

All categories

2 years ago

15

The equation v=F^aL^M^-c were shows the relationship between velocity of the waves tensile force in the string length, L and mas

s, m. Evaluate the value of a, b and d by using dimensional analysis

1 answer:

Travka [436]2 years ago

8 0

I literally looked everywhere for the answer, and I still found nothing. I hope you get it right. Sorry.

You might be interested in

Sound is repeated near hills.why

Anon25 [30]

Answer:

Sound waves are reflected back

5 0

2 years ago

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an

nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0

2 years ago

The plate near the Artic Ridge is among those with the slower growth rates, moving slightly less than 2.5 cm/year. How much will

Aleks04 [339]

Greater than 25 . 2.5 x 10 = 30

4 0

2 years ago

Read 2 more answers

An increase of 180 degrees in Fahrenheit is equal to an increase of how many degrees in celsius?

andriy [413]

Answer:

it is 100 degree which is boiling point in water

7 0

2 years ago

A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 A

Harlamova29_29 [7]

Answer:

(a) Energy density will be equal to $57.31J/m^3$

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area $A=15.9cm^2=15.9\times 10^{-4}m^2$

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H$

(b) Total energy stored in magnetic field $U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J$

(a) Energy density will be equal to

$U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3$

7 0

2 years ago