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3 years ago

15

During resistance exercise, muscles are __________. A. working against a force B. repeatedly contracted C. concerted in movement

D. all of the aboveImproving muscular strength and endurance slows bone density loss.

1 answer:

ExtremeBDS [4]3 years ago

3 0

Answer:

the answer is A. working against a force

Explanation:

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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

When magma cools quickly, what kind of texture or

Ksivusya [100]

Answer:

a

Explanation:

when magma cools Crystal's form because the solution is super saturated with respect to some minerals if the magma cools quickly the crystals do not have much time to form hence they are small and also the resulting rock is fine grained

6 0

3 years ago

Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the

Rzqust [24]

Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

   KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:

   KE = 0.9 PE

(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)

Divide each side by (mass):

   (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

   (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

   Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

      = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

      = (24,500 / 88.2) (m²/s²) / (m/s²)

      = 277-7/9 meters
      (about 911 feet)

3 0

3 years ago

0. 85 kg of lead, specific heat 128 J/kgC, is heated from 50 C to 95 C. How much heat energy did the sample absorb?

Zanzabum

4896

0.85 x 45 x 128 = 4896

Change in energy = specific heat capacity x mass x change in temperature

5 0

2 years ago

A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces

snow_lady [41]

Answer:

<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>

<u><em>Initial velocity = 33 m/s</em></u>

<u><em>Final velocity = 0 m/s</em></u>

<u><em>Average velocity = (33 + 0) / 2 m/s = 16.5 m/s</em></u>

Explanation:

<u><em>First, how long does it take the truck to come to a complete stop?</em></u>

<u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>

<u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>

4 0

3 years ago