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Gnoma [55]
4 years ago
12

You have 13.0 gallons of gasoline. How many liters is this? (1 gallon equals about 3.79 L.)

Physics
2 answers:
Harlamova29_29 [7]4 years ago
8 0

Answer:

I believe the answer is C

Explanation:

Lunna [17]4 years ago
7 0

For this case we must make a conversion, we have as data that:

1 gallon is equivalent to 3.79 liters

By making a rule of three we have:

1 gallon ------------> 3.79 L

13 gallons --------> x

Where "x" represents the equivalent in liters:

x = \frac {13 * 3.79} {1}\\x = 49.27

So, we have 49.27 liters of gasoline.

Answer:

Option C

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The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 hz. Which fre
Natali [406]

Answer:

The frequency 400 hz is not possible .

Explanation:

Given that,

Frequency = 200 hz

Length = l

Suppose, The given frequencies are,

600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.

The possible resonance frequencies are

We need to calculate the fundamental frequency

Using formula of fundamental frequency for pipe

F=\dfrac{nv}{4L}

Where, n = odd number

Put the value of frequency

200= \dfrac{nv}{4L}

We need to calculate the first over tone

Using formula of fundamental frequency

n = 3,

F=\dfrac{nv}{4L}

Put the value into the formula

F_{2}=3\times\dfrac{v}{4l}

F_{2}=3\times200

F_{2}=600\ Hz

We need to calculate the second over tone

Using formula of fundamental frequency

n = 5,

F=\dfrac{nv}{4L}

Put the value into the formula

F_{3}=5\times200

F_{3}=1000\ Hz

We need to calculate the third over tone

Using formula of fundamental frequency

n = 7,

F=\dfrac{nv}{4L}

Put the value into the formula

F_{4}=7\times200

F_{4}=1400\ Hz

We need to calculate the fourth over tone

Using formula of fundamental frequency

n = 9,

F=\dfrac{nv}{4L}

Put the value into the formula

F_{5}=9\times200

F_{5}=1800\ Hz

Hence, The frequency 400 hz is not possible .

8 0
3 years ago
if u connect 3 resistors, having values 2ohm, 3ohm, 5ohm in parallel, will the value of total resistance of will be 2ohm or grea
Tju [1.3M]
<span>If u want only WHether the total resistance is less than 2 or less than 5 or more than 5 ohms:  there is a Simple way.

When you connect resistances in parallel, resultant resistance is always smaller than all of them. So it is less than 2 ohms</span>.

4 0
4 years ago
Read 2 more answers
If the potential difference across the filament is 120 V, what is the strength of the electric field inside the filament
sineoko [7]

Answer:

200 N/C

Explanation:

Given that the length of a 60 W, 240 & Omega light bulb filament is 60 cm. If the potential difference across the filament is 120 V, what is the strength of the electric field inside the filament?

The parameters needed from the question are:

Potential difference V = 120v

Distance ( Length) = 60 cm

Convert cm to m by dividing it by 100

Distance = 60/100

Distance = 0.6 m

The formula for electric field strength E is:

E = V/d

Where

V = potential difference

d = distance

Substitute those parameters in the formula

E = 120 / 0.6

E = 1200 / 6

E = 200 N/C

Therefore, the strength of the electric field inside the filament is 200N/C

5 0
3 years ago
"The velocity of a diver just before hitting the water is -10.0 m/s, where the minus sign indicates that her motion is directly
garri49 [273]

Answer:

Explanation:

Given

velocity of diver u=-10\ m/s i.e. downward motion

acceleration due to gravity a=g=-9.8\ m/s^2

time t=1.16\ s

using equation of motion

y=ut+\frac{1}{2}at^2

y=(-10)\cdot 1.16-\frac{1}{2}(-9.8)(1.16)^2

y=-11.6-6.593

y=-18.19\ m

I.e. in downward direction                    

6 0
3 years ago
the electrical resistance of copper wire varies directly with its length and inversely with the square of the diameter of the wi
leonid [27]

Answer:24.47

Explanation:

Given

L_1=30 m

d_1=3 mm

R_1=25 \Omega

L_2=40 m

d_2=3.5 mm

we know Resistance R=\frac{\rho L}{A}

Where R=resistance

\rho =resistivity

L=Length

A=area of cross-section

A=\frac{\pi d^2}{4}

Thus R\propto \frac{L}{d^2}

therefore

R_1\propto \frac{L_1}{d_1^2}--------1

R_2\propto \frac{L_2}{d_2^2}--------2

divide 1 and 2 we get

\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}

\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}

R_2=25\times 0.734\times 1.33

R_2=24.47 \Omega

8 0
3 years ago
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