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kvv77 [185]
3 years ago
13

Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and

one mass slides on a frictionless 30.0-degree incline. The vertically hanging mass is 6.00 kg and the mass on the incline is 4.00 kg. The acceleration of the 4.00-kg mass is

Physics
1 answer:
lora16 [44]3 years ago
8 0
The tension in the string and the acceleration must be equal for both masses. (See the free body diagrams)





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2.14x 1022 kg 

the 22 is a xponet of ten
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3 years ago
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If a wave has a speed of 12 m/s and a frequency of 4hz what is its wavelength
ELEN [110]
Apply:
wavelength = speed of wave / frequency
wavelength = 12 m/s / 4 Hz = 48 m
6 0
3 years ago
A 3.0 \Omega Ω resistor is connected in parallel to a 6.0 \Omega Ω resistor, and the combination is connected in series with a 4
GarryVolchara [31]

Answer:

The power dissipated in the 3 Ω resistor is P= 5.3watts.

Explanation:

After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.

The resultating resistor is of Req=6Ω.

I= V/Req

I= 2A

the parallel resistors have a potential drop of Vparallel=4 volts.

I(3Ω) = Vparallel/R(3Ω)

I(3Ω)= 1.33A

P= I(3Ω)² * R(3Ω)

P= 5.3 Watts

7 0
4 years ago
Automobile which is being towed at constant velocity up the incline using the cable at c. The automobile has a mass of 5 mg and
Simora [160]

The value of the normal reaction force will be 4.65 kN and 31.76 kN.

<h3>What is force?</h3>

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The normal force on the car is,Na= Nb =1 kN

The force of the cable on the car is,F=1 kN

The force of the gravity is,W=mg

The balancing equation of force;

\rm N_A+N_B+F sin \theta_2 -mg cos \theta_2 =0 \\\\ \rm -Fcos \theta_2 +mg sin \theta_1 =0 \\\\ Fcos \theta_2 \times a -F sin \theta_2 \times b -mg cos \theta_1 \times c -mg sin \theta_1 \times e +N_B (c+d)=0 \\\\ N_A+N_B+F sin 30^0 - 5 \times 9.81 cos 30^0 =0 \\\\ -

Putting the value further;

\rm N_A+N_B+F sin 30^0 - 5 \times 9.81 cos 30^0 =0 \\\\  -F cos 30^0 + 9.81 sin 20^0 =0 \\\\ F cos 30^ 0 (0.6)-F sin30^0 (1.5)-(5 \times 10^6) \times 9.81 \times cos 20^0 (1)-(5)(9.81)sin 20^0 +N_B \times 1.75 =0

Solving the equation, we get;

\rm  F_A = 19.37 \ kN \\\\ N_A=4.65 \ kN \\\\ N_B=31.76 \ kN

Hence, the value of the normal force swill be 4.65 kN and 31.76 kN.

To learn more about the force, refer to the link;

brainly.com/question/26115859

#SPJ4

4 0
2 years ago
A cat walks 1.5km South and then 2.4km East. What is the total displacement of the cat?
Doss [256]

Answer:

Displacement = 2.83km

Explanation:

Given

South = 1.5km

East = 2.4km

Required

Determine the displacement

See attachment for proper representation of the cat's movement.

The displacement is represented by the slanted line and it is calculated as follows:

D\² = S\² + E\²

D^2 = 1.5\² + 2.4\²

D\² = 2.25 + 5.76

D\² = 8.01

D = \sqrt{8.01}

D = 2.83019434

<em>D = 2.83 km (approximated)</em>

8 0
3 years ago
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