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RUDIKE [14]
3 years ago
12

WILL GIVE BRAINLIEST!!!!!

Physics
2 answers:
arsen [322]3 years ago
8 0
I believe it is A, 
a female carrier of the disease.

Hope this helps :)
SOVA2 [1]3 years ago
6 0

Answer:

the answer is letter c.

You might be interested in
The gravitational force of a star on an orbiting planet 1 is F1. Planet 2, which is twice as massive as planet 1 and orbits at t
Andrej [43]

Answer:

ratio = 1 : 4.5

Explanation:

If m₁ is the mass of the star and m₂ the mass of the planet, the force of gravity F₁ for planet 1 is given by:

F_1=\frac{Gm_1m_2}{r^2}

The force F₂:

F_2=\frac{Gm_1(2m_2)}{(3r)^2}

The ratio:

\frac{F_2}{F_1}=\frac{2}{9}

8 0
3 years ago
A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 17.3 m/s directly upward.
elena-s [515]

Answer:

h=15.27m

Explanation:

Since at maximum height the vertical velocity must be null it's better to use the formula:

v_f^2=v_i^2+2ad

We will use this formula for the vertical direction, choosing the upward direction as the positive one, so we have:

0=v_i^2+2ah

or

h=-\frac{v_i^2}{2a}

which for our values is:

h=-\frac{(17.3m/s)^2}{2(-9.8m/s^2)}=15.27m

7 0
3 years ago
The density of air is 1.3 kg/m^3. What mass of air is contained in a room measuring 2.5m x 4m x 10m? Give your answer in kg. Bra
krek1111 [17]
Mass= density x volume
1.3 kg/m^3 x ( 2.5x4x10) m^3
= 130 kg
7 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_aP_{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
2 years ago
If an object's velocity changes from 25 meters per second to
damaskus [11]

Answer:

B

Explanation:

the way to get b you have divion each other

3 0
2 years ago
Read 2 more answers
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