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3 years ago

You can use one of your 5 senses to make _____ during an inquiry activity.

Physics

2 answers:

AURORKA [14]3 years ago

6 0

the answer is noise.

Not nose, noise.

aivan3 [116]3 years ago

4 0

you can use one of your 5 senses to make _____ during an inquiry activity.

sound

You might be interested in

If you want the ball to land in your hand when it comes back down, should you toss the ball straight upward, in a forward direct

denpristay [2]

Straight upward

the ball moves in the forward direction with your walking speed at all times. If you want the ball to land in your hand when it comes back down, you should toss the ball straight upward.

<h3>What is Projectile motion ?</h3>

Projectile motion is the motion of an object thrown (projected) into the air.

After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory

A projectile can be a thrown ball, a bullet or a springboard diver ... Except for air resistance, the forward velocity of any projectile is constant and is equal to the initial velocity when it was released.

Learn more about Projectile motion here:

brainly.com/question/27116954

#SPJ4

7 0

1 year ago

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

<u>At State 1:</u>

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0

3 years ago

As a way of determining the inductance of a coil used in a research project, a student first connects the coil to a 5.62 V batte

Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

5 0

3 years ago

What are loops of gas on sun that link different parts of sunspot regions together?

love history [14]

So we want to know what are loops of gas on the Sun that link different parts of sunspot regions together. A large and bright gaseous feature that extends from the surface of the Sun that links different parts of sunspot regions together is called Prominence. They are on the Suns surface in the photosphere and they extend outwards into the Corona.

7 0

3 years ago

What is an ideal diode? a. The ideal diode acts as an open circuit for forward currents and as a short circuit with reverse volt

Serhud [2]

Answer:

d. The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applied.

Explanation:

Ideal diode acts like an ideal conductor. In case of forward voltage it acts like an ideal conductor. However when it is reverse biased then it behaves like an ideal insulator. You can understand it bu considering a switch. When the voltage is forward then ideal diode acts like a closed switch. When the voltage is reverse biased then ideal diode behaves like an open switch.

That is why we can say that the ideal diode acts as a short circuit (higher conduction) for forward currents and as an open circuit ( zero conduction) with reverse voltage applied.

8 0

3 years ago