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3 years ago

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After a traffic light turns green, a car is observed to accelerate from rest for 6.0 seconds. If the speed limit on that stretch

of road is 24 m/s, what is the maximum acceleration the car could have without breaking the law? (Show your work or explain your reasoning)

1 answer:

nataly862011 [7]3 years ago

8 0

congratulations oh feedcounsel Ozark Ozark

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John stretched while his muscles were cold. Which negative effect could occur?

Genrish500 [490]

<h3><u>Answer;</u></h3>

C) He could injure or pull a muscle.

<h3><u>Explanation</u>;</h3>

<em><u>When muscles are stretched the muscle fibers are temporary lengthened. Muscles have a unique ability to undergo contraction and lengthening since they are elastic.</u></em>
<em><u>When the muscles are warm they are more elastic, therefore, warming up the muscles is very important for the purpose of avoiding injuries.</u></em> Ir is important to engage in moderate cardiovascular warm-up prior to stretching, which increases the blood flow to the active and thus avoiding injuries to the muscles.
Stretching muscles when they are cold may bring injuries or cause muscle pulls, which are painful.

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3 years ago

Read 2 more answers

What is an example of a mechanical noise?

faltersainse [42]

Resonance:
The resounding recurrence is the recurrence at which a bit of metal, plastic or whatever else swings/vibrates with minimal measure of vitality input. Think about a man on a play area swing. You realize that it requires next to no push to keep the individual swinging. The recurrence at which they swing forward and backward is their full recurrence. In the event that you endeavor to influence them to swing speedier or slower, it will take altogether more vitality.

Resonating Panels:
This kind of clamor is caused when the bass notes are an indistinguishable recurrence from the thunderous recurrence of a metal or plastic board. To stop or decrease the commotion related with this kind of issue, you can do two or three things.

Rattling:
This sort of commotion would be caused when 2 bits of metal, plastic, whatever... are sufficiently close to hammer into each other when they resound. This is most likely best illuminated by filling the hole between the two vibrating parts with silicone sealant or shut cell froth climate stripping. The climate stripping is a superior arrangement in places like behind the tag. On the off chance that you have a tag outline, you can get some truly thin climate stripping and put between the casing and the plate.

4 0

3 years 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

A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?

Margarita [4]

Answer:

The answer to the question is 7200

7 0

3 years ago

Make a rough estimate of the number of quanta emitted in one second by a 100 W light bulb. Assume that the typical wavelength em

mixas84 [53]

Answer:

#_photon = 5 10²⁰ photons / s

Explanation:

For this exercise let's calculate the energy of a single quantum of energy, use Planck's law

E = h f

c= λ f

E = h c / λ

λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m

Let's calculate

E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹

E₀ = 19.89 10⁻²⁰ J

This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w

#_photon = P / E₀

#_photon = 100 / 19.89 10⁻²⁰

#_photon = 5 10²⁰ photons / s

6 0

3 years ago