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Nady [450]
3 years ago
15

Find the equivalent resistance in the circuit between A and B ​

Physics
1 answer:
Monica [59]3 years ago
5 0

Answer:

6Ω

Explanation:

The 6Ω resistor and the 3Ω resistor are in parallel.

R = 1 / (1/6 + 1/3)

R = 1 / (1/6 + 2/6)

R = 1 / (3/6)

R = 6/3

R = 2

The 8Ω resistor and the other 8Ω resistor are also in parallel.

R = 1 / (1/8 + 1/8)

R = 1 / (2/8)

R = 8/2

R = 4

This 4Ω resistance is in series with the 4Ω resistor.

R = 4 + 4

R = 8

The 2Ω resistor and the 6Ω resistor are also in series.

R = 2 + 6

R = 8

These two 8Ω resistances are in parallel.

R = 1 / (1/8 + 1/8)

R = 1 / (2/8)

R = 8/2

R = 4

Finally, this 4Ω resistance is in series with the 2Ω resistance we found earlier.

R = 4 + 2

R = 6

The total equivalent resistance is 6Ω.

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A satellite orbiting above the earth needs no power source to keep orbiting the earth.
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Read 2 more answers
The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?
Ivanshal [37]
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration g=9.81 m/s^2. Calling h its height at t=0, the height at time t is given by
h(t)=h- \frac{1}{2}gt^2
We are told thatn when t=0.75 s the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
0=h- \frac{1}{2}gt^2
h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(0.75 s)^2=2.76 m

2) The speed of the grapefruit at time t is given by
v(t)=v_0 +gt
where v_0=0 is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
v(0.75 s)=gt=(9.81 m/s^2)(0.75 s)=7.36 m/s
3 0
4 years ago
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