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Liono4ka [1.6K]
3 years ago
11

You are pushing on a heavy door, trying to slide it open. If your friend stands behind you and helps you push, how have the forc

es changed?
a. The net force applied will increase.
b. The net force will decrease.
c. There will be less friction.
d. The applied force does not change.
Physics
2 answers:
Art [367]3 years ago
8 0
A...... The net force applied will increase.
ad-work [718]3 years ago
7 0

Answer: A) The net force applied will increase.

Explanation:

Both of us trying to push the door will increase the net force and this is because the force is been applied by both of us in the same direction. Forces acting in the same direction are added up to produce a resultant force which will cause an increment in the total force.

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4 years ago
A 0.18 m radius pulley is free to rotate about a horizontal axis. A mass and a mass are attached by a massless string, which is
Musya8 [376]

Answer:

T = 1.766(M-m) Nm where M and m are the 2 masses of the objects

Explanation:

Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.

Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m

T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm

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3 years ago
Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and
snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the  y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be v_{P/A} = v( cos \theta \ i + sin \theta \ j)

where:

v_{P/A} = velocity of the plane in regard to the air

v = velocity

θ =  angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

v_{P} = v_o( cos \phi \ i + sin \phi \ j)

where;

v_p = velocity of the airplane

v_o = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing  v_o  = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

v_p = v_A + v_{P/A}

v_A=  v_P  -v_{P/A}   --- (1)

v_A= ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

v_A= (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

v_A= (-39.24 km/h)i + (13.44 km/h) j

|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}

v_A = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ  = tan ^{-1} (2.9)

θ  =  70.97°

The angle of motion is  70.97° from west of north with a velocity of 41.48 km/h.

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Answer:

a

Explanation:

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