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3 years ago

Can anyone help me with any parts in question number 1: parallel circuit?

Physics

1 answer:

Leno4ka [110]3 years ago

4 0

Question #1:

a). The sketch is attached to this answer.

b). The equivalent resistance of 30Ω and 50Ω in parallel is

1 / (1/30 + 1/50) = 18.75 Ω

c). I = V/R = (100/30) = (3 and 1/3) Amperes

d). Follow the wires, and you see that the 50Ω resistor is
connected directly to the battery, and so is the voltmeter.
So the voltage across the 50Ω resistor, and the reading
on the voltmeter, is 100 volts.

e). I = V/R
Through the 30Ω resistor: I = 3-1/3 A
Through the 50Ω resistor: I = 2 A

f). In the parallel circuit, both resistors are connected
directly to the battery. So neither resistor even knows
that the other one is there. Each resistor sees 100 volts,
and the current through each resistor is 100/R, just as if
it were the only resistor in the circuit.

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A small bead with a positive charge q is free to slide on a horizontal wire of length 4.5 cm . At the left end of the wire is a

-Dominant- [34]

Answer:

1.5cm

Explanation:

The place where the bead will come to rest is the place where the force between the left charge and the bead is the same, but in opposite direction, as he force between the right charge and the bead.

$F_{left} = F_{right}\\K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}$

Also:

$r_{left}+r_{right}=0.045m\\r_{right} = 0.045m-r_{left}$

$q_{left}= q\\q_{right}=4q$

So, we replace and simplify. Notice that the charges and the Coulomb constant will cancel:

$K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}\\\frac{q*q}{r_{left}^2} = \frac{4q*q}{(0.045m-r_{left})^2}\\(0.045m-r_{left})^2 =4r_{left}^2\\0.002025m - 2*0.045r_{left} + r_{left}^2 = 4r_{left}^2\\3r_{left}^2 + 0.09r_{left} - 0.002025m = 0$

$r_{left} = \frac{-(0.09)+-\sqrt{(0.09)^2-4*(3)(-0.002025)}}{2(3)}\\ r_{left} = 0.015m| -0.045m$

But the only solution that would place the bead on the wire is 0.015m or 1.5cm, so this is our answer.

5 0

3 years ago

You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc

pickupchik [31]

a) $15.4^{\circ}$

b) 5.2 m/s

c) 151.2 N

Explanation:

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

$mg sin \theta$

where

m =16 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

$\theta$ is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

$F=ma\\mgsin \theta = ma$

where a is the acceleration.

From the equation above we get

$a=g sin \theta$

And we are told that the acceleration must not exceed

$a=2.6 m/s^2$

Substituting this value and solving for $\theta$ , we find the maximum angle of the ramp:

$\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}$

Here we are told that the vertical distance of the ramp is

$h=1.4 m$

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

$GPE=KE\\mgh=\frac{1}{2}mv^2$

where:

m = 16 kg is the mass of the box

$g=9.8 m/s^2$

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s$

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

$mg cos \theta$

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

$N-mg cos \theta$

and by substituting:

m = 16 kg

$g=9.8 m/s^2$

$\theta=15.4^{\circ}$

We can find the normal force:

$N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N$

8 0

3 years ago

How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800

ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

$A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2$

The inductance of the solenoid is given by;

$L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H$

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0

3 years ago

If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is

Law Incorporation [45]

Answer:

Explanation:

v = v⁰ (its original speed) + a (negative acceleration) X t² (time)

v = 15 - 10 x 1.2 = 15 - 12 = 3 (it's slowing down)

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2 years ago

Question 1

mestny [16]

Answer:

false

A hypothesis states a presumed relationship between two variables in a way that can be tested with empirical data. ... The cause is called the independent variable; and the effect is called the dependent variable.

6 0

3 years ago