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Luba_88 [7]
3 years ago
10

Two resistors are to be combined to form an equivalent resistance of 1000 Ω. The resistors are taken from available stock on han

d as acquired over the years. Readily available are two common resistors rated at 500 ± 50 Ω and two common resistors rated at 2000 ± 100 Ω. (a) Calculate the uncertainties for both combinations. (b) What combination of resistors (series or parallel) would provide the smaller uncertainty in an equivalent 1000 Ω resistance?

Physics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

Even though each of 2000 ohms resistor has twice the uncertainty of a 500 ohms resistor, the parallel circuit has half the uncertainty of the series circuit

Thus, the PARALLEL CIRCUIT would be more accurate

Explanation:

Check attachment for solution

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Uranium (atomic number 92)
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A lad, waiting for his friend walks in the sidewalk, in front of her house, from the front door, first, he moves towards the Pos
Andreas93 [3]

His total displacement from his original position is -1 m

We know that total displacement of an object from a position x to a position x', d = final position - initial position.

d = x' - x

If we assume the lad's initial position in front of her house is x = 0 m. The lad then moves towards the positive x-axis, 5 m. He then ends up at x' = 5 m. He then finally goes back 6 m.

Since displacement = final position - initial position, and his displacement is d' = -6 m (since he moves in the negative x - direction or moves back) from his initial position of x' = 5 m.

His final position, x" after moving back 6 m is gotten from

x" - x' = -6 m

x" = -6 + x'

x" = -6 + 5

x" = -1 m

Thus, his total displacement from his original position is

d = final position - initial position

d = x" - x

d = -1 m - 0 m

d = -1 m

So, his total displacement from his original position is -1 m

Learn more about displacement here:

brainly.com/question/17587058

3 0
3 years ago
A basketball rolls without slipping (starting from rest) down a ramp. If the ramp is sloped by an angle of 4 degrees above the h
slavikrds [6]

Answer:

11.7 m/s

Explanation:

To find its speed, we first find the acceleration of the center of mass of a rolling object is given by

a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²

a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)

  = 9.8 m/s²sin4(1 + 2/3)

  = 9.8 m/s²sin4 × (5/3)

  = 1.14 m/s²

To find its speed v after rolling for 60 m, we use

v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m

v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s

4 0
3 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
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