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2 years ago

Name the force acting on the handles of a plastic shopping bag that is full of groceries.

Engineering

1 answer:

Dimas [21]2 years ago

3 0

The answer is

The force acting on the handles of a plastic shopping bag that is full of groceries are gravitational force, muscular force.

Gravitational force works downwards, and muscular force is applied by the hands lifting bucket upwards. Two forces are in the opposite direction and hence balancing each other. As a result net force is zero.

Learn more things about gravitational and muscular force

brainly.com/question/20939728

You might be interested in

Wattage is:

Ksju [112]

Answer:

c.Both A and B.

Explanation:

the wattage is c and d

7 0

2 years ago

The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc

a_sh-v [17]

Answer:

(a)

dQ = mdq

dq = $C_p$ dT

$q = \int\limits^{T_2}_{T_1} {C_p} \, dT$ = $C_p$ (T₂ - T₁)

From the above equations, the underlying assumption is that $C_p$ remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ , Q₂ = 6.14 KJ

ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter

Let $C_{cal}$ be heat constant of calorimeter

Q₂ = $C_{cal}$ ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m $C_p$ ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³ at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m $C_p$ ' ΔT

$C_p$ = (16.73 - 6.14) / (15.085 x 3.10)

$C_p$ = 0.22646 KJ mol⁻¹ k⁻¹

6 0

2 years ago

5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h

frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

V2max = √2 * 9.81 * 3

V2max = √58.86 = 7.67 m/s

maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

Qmax = 0.00071 * 7.67 = 0.00545 m³/s

Qmax = 5.45 L/s

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Actual flow rate will be less because of heads such as friction head and pressure head.

7 0

3 years ago

A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar

Westkost [7]

Answer:

a) $m_{2} = 0.753\,kg$ , b) $Q_{in} = 2122.963\,kJ$

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.02726\,\frac{m^{3}}{kg}$

$u = 1192.94\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.2$

State 2 - Gas-Vapor Mixture

$P = 1500\,kPa$

$T = 198.29^{\textdegree}C$

$\nu = 0.06643\,\frac{m^{3}}{kg}$

$u = 1718.12\,\frac{kJ}{kg}$

$h_{g} = 2791.0\,\frac{kJ}{kg}$

$x = 0.5$

The model for the rigid tank is created by using the First Law of Thermodynamics:

$Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}$

Initial and final masses are:

$m_{1} = \frac{V_{1}}{\nu_{1}}$

$m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }$

$m_{1} = 1.834\,kg$

$m_{2} = \frac{V_{2}}{\nu_{2}}$

$m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }$

$m_{2} = 0.753\,kg$

a) The final mass within the tank is:

$m_{2} = 0.753\,kg$

b) The total amount of heat transfer is:

$Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}$

$Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )$

$Q_{in} = 2122.963\,kJ$

5 0

3 years ago

The bulk modulus of a fluid if it undergoes a 1% change in volume when subjected to a pressure change of 10,000 psi is (a) 0.01

Veseljchak [2.6K]

Answer:

The required bulk modulus is $10^{6}$ Psi. So, the answer is non of these.

Explanation:

Change in pressure of the fluid is directly proportional to the volumetric strain. The constant of proportionality is the bulk modulus of the fluid.

Step1

Given:

Percentage change in volume is 1%.

Change in pressure is 10000 Psi.

Calculation:

Step2

Volumetric strain is calculated as follows:

$\frac{\bigtriangleup V}{V}=\frac{1}{100}$

$\frac{\bigtriangleup V}{V}=0.01$

Step3

Bulk modulus is calculated as follows:

\frac{\bigtriangleup V}{V}=0.01

$\frac{\bigtriangleup V}{V}=0.01$

$10000=K\times0.01$

K = 1000000 Psi.

Thus, the required bulk modulus is $10^{6}$ Psi.

3 0

3 years ago