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egoroff_w [7]
3 years ago
15

What is the definition of balanced forces? i need help i cnt find the definition on dictionary.com

Physics
1 answer:
kicyunya [14]3 years ago
8 0
Any group of 2 or more forces is balanced if the sum of all of their strengths and directions adds up to zero. Here's an example that I just made up:. Take 50 ropes. Tie one end of every rope all together into one big fat knot, and then have a tug of war in 50 different directions. Get 10 football players and 40 cheerleaders to each take the end of one rope, then everybody spread out and start pulling. It's possible that if everybody stood in just the right place, they might all cancel each other out, and the big knot in the center might just hang there and not move at all ... just as if there was NO FORCE on it at all. If that happened, we would know that the group of 50 forces acting on the knot is "balanced".
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1 (KE, PE) An object of mass 90 lbm is projected straight upward from the surface of the earth and reaches a height of 900 ft wh
Natali5045456 [20]

Answer:

Initial kinetic energy will be 2608200 lbf-ft

Initial speed will be 240.7488 m/sec                

Explanation:

We have given height h = 900 ft

Mass of the object m = 90 lbm

Acceleration due to gravity g=32.2ft/sec^2

Initial kinetic energy will be equal to potential energy of the object

So kinetic energy = potential energy = mgh =90\times 32.2\times 900=2608200\ lbf-ft

This energy is also equal to the initial kinetic energy

So \frac{1}{2}mv^2=2608200

v=240.7488m/sec

So initial velocity will be 240.7488 m/sec

7 0
4 years ago
What is the answer to these questions?
Oksana_A [137]

Answer:

THES IS NOT

Explanation:

THIS PAPAER IS A FAKE PAPAER BEACISE POGI TALAGA AKO

7 0
3 years ago
Science in the study of the natural world through. And.
Lyrx [107]
You need to finish the question
6 0
3 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

mg sin \theta - \mu N = ma (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, \theta=39^{\circ} being the angle of the incline

\mu N is the frictional force, with \mu = 0.6 being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

N-mg cos \theta =0

where mg cos \theta is the component of the weight in the direction perpendicular to the slope. Solving for N,

N=mg cos \theta

Substituting into (1), solving for a, we find the acceleration:

a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2

6 0
4 years ago
Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the
Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +delta-p

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +0

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

0-0=h_{2} -400x10^{3}  +\frac{40^{2}-250^{2}      }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}

The exit area is:

m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}

6 0
3 years ago
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