All categories

2 years ago

Can a body possess velocity at the same time in horizontal and vertical directions?

Physics

1 answer:

iogann1982 [59]2 years ago

5 0

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

You might be interested in

A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f

zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0

3 years ago

Read 2 more answers

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

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2. A fan is 30 m from the loudspeak

aliina [53]

Explanation:

Below is an attachment containing the solution.

5 0

3 years ago

3.1

Ilia_Sergeevich [38]

Answer:

The value is $C = 30729\ c$

Explanation:

From the question we are told that

The power rating of the stove is $P = 4.4 KW = 4.4 *10^{3} \ W$

The duration of its use everyday is $t_1 = 70 \ minutes = 1.167 \ hours$

The rating of the light bulbs is $P_2' = 150 W$

The number is $n = 7$

The power rating of the total bulb is $P_2 = 7 * 150 = 1050 \ W$

The duration of its use everyday is $t_2 = 7 hours$

The power rating of miscellaneous appliance $P_3 = 1.8 \ KW = 1.8 *10^{3} \ W$

The duration of its use everyday is $t_3 = 1 hour$

The power rating of hot water $P = 4 KW = 4 *10^{3} \ W$

The duration of its use everyday is $t_4 = 120 \ minutes = 2 hours$

Generally the total electrical energy used in 1 month is mathematically represented as

$E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30$

=> $E = 4.4*10^{3} * 1.167 * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30$

=> $E = 614598 \ W \cdot h$

=> $E = 614.6 \ K W \cdot h$

Generally the monthly electricity bill is mathematically represented as

$C = 614.6 * 50$

=> $C = 30729\ c$

8 0

2 years ago

Kevin is a major stockholder in Professional Transmission Services (PTS), a nationwide network of transmission repair shops foun

const2013 [10]

Answer:

Kevin wants to make the firm private

Explanation:

Based on the information provided within the question in regards to the situation it seems that Kevin wants to make the firm private. Private Firms are companies that are owned by non-governmental entities or instead owned by a single individual or a very small amount of shareholders. Which is what Kevin seems to be wanting to do since he wants to buy back all the shares of the company so that only his family owns the company.

If you have any more questions feel free to ask away at Brainly.

6 0

3 years ago

Can a body possess velocity at the same time in horizontal and vertical directions?​

Can a body possess velocity at the same time in horizontal and vertical directions?