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3 years ago

Which is not expecting acceleration?

Physics

1 answer:

Elena L [17]3 years ago

4 0

A point on the outside of a spinning top whose rotational speed is constant.

Since the point at the top is at rest so it will not accelerate at all

A skydiver whose air resistance is equal to that of her weight.

Since weight is counter balanced by air resistance so net force on it is zero so acceleration will be zero

A car on the freeway experiencing a net force of -120 N.

Since net force is not zero so it will accelerate here

A submerged beach ball whose buoyant force is eight times the force of gravity on it.

Since buoyant force is more than the weight so it will not at rest and move upwards with some acceleration

A frictionless spinning merry go round.

Since spinning wheel accelerating due to centripetal force so it is non inertial frame

A falling rock.

Falling rock is accelerating due to gravity so it is again non inertial frame

A hot air balloon moving at 30 degrees east of north with no net force.

since hot air balloon is having no force on it so its acceleration is zero and its an inertial frame

A space shuttle whose boosters just ignited for takeoff.

Since space shuttle is accelerating due to boosters so it is non inertial frame

Here net force on it is due to two given forces

$F_1 = 4320 N$ towards right

$F_2 = 4380 N$ towards left

so here net force is given as

$F = F_2 - F_1$

$F = 4380 - 4320 = 60 N$

since force towards left is more so the direction will be towards left

so correct answer is 60 N towards left

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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

At State 1:

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:

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

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

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

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

At State 1:

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.

At State 1;

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

At State 2:

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

The period of a 261-Hertz sound wave is

SCORPION-xisa [38]

The period of a 261 hertz sound wave is
.000383 seconds

7 0

3 years ago

49 = 1/2gt²

BaLLatris [955]

Answer:

e son

Explanation:

se lo on

7 0

3 years ago

Which color is in the middle of the light spectrum?

Elanso [62]

Well, with the light spectrum there technically is no middle color. Both green and yellow meet up in the middle at 560 nm (wavelength interval) and 540 THz (frequency interval).

3 0

2 years ago

AHH please help-

umka2103 [35]

Answer:

Radiation

Explanation:

I hope that help's

5 0

3 years ago