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3 years ago

8

Round your answers to one decimal place. This parallel circuit has two resistors at 15 and 40 ohms. What is the total resistance

? Ω If there is a 12 V battery in the circuit, what would the approximate current be? A

2 answers:

Dmitriy789 [7]3 years ago

7 0

Answer:

10.9 ohms .

1.1 Amperes

Explanation:

goldenfox [79]3 years ago

4 0

-- With two resistors in parallel, the total effective resistance is
the reciprocal of (1/R₁ + 1/R₂).

1/R₁ + 1/R₂ = 1/15 + 1/40

= 8/120 + 3/120

= 11/120

So the total effective resistance is 120/11 = 10.9 ohms .

Current = (voltage) / (resistance)

= 12 / (120/11)

= (12 · 11) / 120

= 132/120 = 1.1 Amperes

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A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

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Answer:

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Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

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2 years ago

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Stress. Being alone. having nothing

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3 years ago

An important dimensionless parameter in certain types of fluid flow problems is the Froude number defined as V/√g.l where V is a

prohojiy [21]

Answer:

1.24611

Explanation:

V = Velocity = 10 ft/s

L = Length = 2 ft

g = Acceleration due to gravity = 32.2 ft/s²

Froude number is given by

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10}{\sqrt{32.2\times 2}}\\\Rightarrow Fr=1.24611$

Converting to SI units

$10\ ft/s=10\times \dfrac{1}{3.281}$

$32.2\ ft/s^2=32.2\times \dfrac{1}{3.281}$

$2\ ft=2\times \dfrac{1}{3.281}$

$Fr=\dfrac{V}{\sqrt{gL}}\\\Rightarrow Fr=\dfrac{10\times \dfrac{1}{3.281}}{\sqrt{32.2\times \dfrac{1}{3.281}\times 2\times \dfrac{1}{3.281}}}\\\Rightarrow Fr=1.24611$

The Froude number is 1.24611

The Froude number is equal. The Froude number is dimensionless as the units cancel each other. In order for this to happen the units used need to be consitent either imperial or SI.

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3 years ago

The type of energy that depends on position is called

attashe74 [19]

The type of energy that depends on position is called
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3 years ago

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the

DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 $m/s^2$ (block), aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got N = mg cos θ,

f = $\mu_s N$

= $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$ .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$ ................(v)

Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

6 0

2 years ago