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3 years ago

Which of the following are not conditions for producing work?

Physics

2 answers:

jeka943 years ago

8 0

Answer:

A and C

Explanation:

We know that

Work done,W=Fd

Where F=Applied force

d=Displacement of object

Momentum, p= mv

It is not necessary that momentum must be greater than force .If momentum is less than force then the work can be produced because work depend upon displacement and applied force.

$W=Fdcos\theta$

If the value of theta is not equal to zero the work can be produced if theta is equal to 30 ,45 or 60 degree.

Therefore, it is not necessary that atleast part of the applies force must act along the same line object moves.

If force is greater than weight then, the work can be produced .

If force is less than the weight then, the work cannot be produced .

Hence, the force must be very large.

TEA [102]3 years ago

6 0

Answer:D

Explanation:

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den301095 [7]

Hello there,

It takes 300 newtons of force and a distance of 20 meters for a moving cart to come to a stop. How much kinetic energy did this cart have?

Answer: 6000

8 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its

m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0

3 years ago

The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro

Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

Pressure ,P = 1 atm

As we know that psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

Specific humidity,ω = 0.00922 kg/kg

The enthalpy ( h)

h=47.59 KJ/kg

The relative humidity, RH

RH= 49.58 %

Specific volume ,

v= 0.853 m³/kg

5 0

2 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

2 years ago

In the following lever system, what is the clockwise torque? 18 Nm 240 Nm 360 Nm 480 Nm

MaRussiya [10]

Answer:

240 Nm

Explanation:

The clockwise torque is the torque determined only by the force that makes the lever rotating clockwise: therefore, the force of 80 N on the right.

The torque produced by this force is given by:

$\tau = F d$

where

F is the magnitude of the force

d is the arm

For the force of 80 N on the right,

F = 80 N

d = 3 m (distance from the pivot)

So, the clockwise moment is

$\tau = (80)(3)=240 Nm$

3 0

3 years ago

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