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SVETLANKA909090 [29]

3 years ago

15

Please Help

2 answers:

satela [25.4K]3 years ago

4 0

A man pulls on a box with 50 N of applied force while another man pushes on the box from the opposite direction with 60 N of applied force. The box will stand still.

TRUE.

barxatty [35]3 years ago

4 0

Answer:

False statement.

Explanation:

It is given that,

Force with which a man pulls a box, F = 50 N (one side)

Force with which pushes on the box, F' = 60 N (in opposite direction)

The forces acting on the box are unbalanced. It means that the force acting in one side is more or less then force in other side. As a result the box will move.

So, the given statement is false.

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â/8.37 points scalcet8 12.2.037. ask your teacher my notes question part points submissions used a block-and-tackle pulley hoist

AnnyKZ [126]

Refer to the diagram shown below.

The hoist is in static equilibrium supported by tensions in the two ropes.

For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)

For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)

Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979

Answer:
T₂ = 225.12 N
T₃ = 275.98 N

4 0

3 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago

Non examples of gravity ?

Afina-wow [57]

Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.

3 0

2 years ago

Read 2 more answers

Nellie newton swings a rock into a circular path while holding an attached string overhead. the string makes a 45-degree angle t

taurus [48]

Horizontal component of the string tension

6 0

2 years ago

PLEASE PLEASE HELPPPPPP!!!!!!!’

Ostrovityanka [42]

The answer is B

explanation

3 0

2 years ago