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2 years ago

Please Someone Help Me...

Physics

1 answer:

goldenfox [79]2 years ago

6 0

Answer:

Hi myself Shrushtee.

Explanation:

The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current. Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire.

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A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas

leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₁² v₂

$v_1= \dfrac{r_1^2}{r_2^2} v_2$

$v_1= \dfrac{1.4^2}{2.8^2} v_2$

$v_1= 0.25 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)$

$\Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)$

$v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}$

$v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}$

v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)² x 1.97

Q = 1.21 x 10⁻³ m³/s

5 0

3 years ago

An empty glass beaker has a mass of 103 g. When filled with water, it has a total mass of 361g.

aivan3 [116]

Answer:

0.96 gcm¯³

Explanation:

From the question given above, the following data were obtained:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of beaker + oil = 351 g

Density of water = 1 gcm¯³

Density of cooking oil =?

Next, we shall determine the mass of water. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + water = 361 g

Mass of water =?

Mass of water = (Mass of beaker + water) – (Mass of empty beaker)

Mass of water = 361 – 103

Mass of water = 258 g

Next, we shall determine the volume of the beaker. This can be obtained by calculating the volume of water in the beaker.

Density of water = 1 gcm¯³

Mass of water = 258 g

Volume of water =?

Density = mass /volume

1 = 258 / volume

Cross multiply

1 × volume = 258

Volume of water = 258 cm³

Thus the volume of the beaker is 258 cm³.

Next, we shall determine the mass of the cooking oil. This can be obtained as follow:

Mass of empty beaker = 103 g

Mass of beaker + oil = 351 g

Mass of cooking oil =?

Mass of cooking oil = (Mass of beaker + oil) – (Mass of empty beaker)

Mass of cooking oil = 351 – 103

Mass of cooking oil = 248 g

Finally, we shall determine the density of the cooking oil. This can be obtained as follow:

Mass of cooking oil = 248 g

Volume of the beaker = 258 cm³

Density of cooking oil =?

Density = mass / volume

Density = 248 / 258

Density of cooking oil = 0.96 gcm¯³

7 0

2 years ago

Buttery popcorn contained in a large 1__bowl has a mass of about 50 __<br>and about 650 calories.

Rasek [7]

Answer:

Buttery popcorn contained in a large 1 liter bowl has a mass of about 50 mg and about 650 calories.

Explanation:

Liter is the most appropriate unit to measure a bowl. Usually 1 liter of liquid has a mass of 1000 gram. Since popcorn is something lightweight and only a few can fill the bowl quickly so 50 mg makes perfect sense with 1 liter of bowl and 650 calories in buttery popcorn.

4 0

3 years ago

Read 2 more answers

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$