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3 years ago

Which statement about energy transformations is not true?

Physics

1 answer:

frutty [35]3 years ago

3 0

Answer:

Explanation:

Let's consider the conservation of energy

Energy can not be destroyed or created energy can be transferred one form energy to a another form of energy.

(A) Then all form of energy can be transfer into any form of energy . There for all forms of energy can be transformed to thermal energy

(B) This is the law of conservation of energy

(C)Potential chemical energy can be transferred into any form of energy. It's not only to kinetic energy

(D) Some amount of thermal energy always released when it's transferred energy due to reaction forces or frictional forces.

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How would you make a structure stronger?

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Make the foundation more sturdy

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4 years ago

LuLu mixes a solution of vinegar for a homemade salad dressing she is making. She makes 300 mL of vinegar by mixing 135 mL aceti

photoshop1234 [79]

Answer:

percentage by volume of the solute = 45%

Explanation:

The percentage by volume of the solute in a solution is the percentage of the volume of the solute as against the overall volume of the solution.

In this example,

volume of solution = 300mL

volume of solute = 135mL

% volume of solute = (volume of solute ÷ volume of solution) × 100

= (135 ÷ 300) × 100

= 0.45 × 100 = 45%

∴ percentage by volume of the solute = 45%

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3 years ago

Two loudspeakers on a concert stage are vibrating in phase. A listener 50.5 m from the left speaker and 26.0 m from the right on

mixer [17]

To solve this problem, it is necessary to apply the concepts related to the constructive interference caused by the wavelengths of the sound traveling in the air.

From the definition of constructive interference we know that

$\Delta x = n \lambda \rightarrow \lambda = \frac{\Delta x}{n}$

Where

$\Delta x =$ Distance between speakers

n = Integer which represent he number of repetition of the spectrum

$\lambda =$ Wavelength

At the same time the frequency is subject to the form,

$f = \frac{v}{\lambda}$

Where

$v = 343m/s \rightarrow$ Velocity (of the sound at this case)

From our given values we have to $\Delta x$ is

$\Delta x = 50.5m-26m$

$\Delta x = 24.5m$

The wavelength would be subject to the sound spectrum therefore for n = 1,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{1}$

$\lambda = 24.5m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{24.5}$

$f = 14Hz$

For the value of n = 2,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{2}$

$\lambda = 12.25m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{12.25}$

$f = 28Hz$

For n = 3,

$\lambda = \frac{\Delta x}{n}$

$\lambda = \frac{24.5}{3}$

$\lambda = 8.167m$

Then the frequency would be,

$f = \frac{v}{\lambda}$

$f = \frac{343}{12.25}$

$f = 42Hz$

From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz

5 0

3 years ago

Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of launch is 75.0° and 15.0° resp

stepladder [879]

The answer is the range of A and B is equal

7 0

3 years ago

Read 2 more answers

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

3 years ago