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2 years ago

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An aluminum wire of length 1.0 meter has a resistance of 9.0x10^-3 ohm. If the wire were cut into two equal lengths, each length

would have a resistance of

2 answers:

murzikaleks [220]2 years ago

0 0

Answer:since resistance depends on length, then the resistance will be shared equally since it was cut equally therefore the resistance is

4.5*10^-3 ohm

Explanation:

Diano4ka-milaya [45]2 years ago

0 0

Answer : The each length would have a resistance of $4.5\times 10^{-3}\Omega$

Explanation :

Let the resistance of equal length of wire be, r

That mean,

$r+r=R$

where,

r = resistance of equal length of wire

R = total resistance of wire = $9.0\times 10^{-3}\Omega$

As we know that resistance is directly proportional to length.

Total resistance of aluminum wire = Sum of resistance of two equal piece of wires

$r+r=9.0\times 10^{-3}\Omega$

$2r=9.0\times 10^{-3}\Omega$

$r=\frac{9.0\times 10^{-3}\Omega}{2}$

$r=4.5\times 10^{-3}\Omega$

Therefore, the each length would have a resistance of $4.5\times 10^{-3}\Omega$

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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago

A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s

BlackZzzverrR [31]

Answer:

The speed in the first point is: 4.98m/s

The acceleration is: 1.67m/s^2

The prior distance from the first point is: 7.42m

Explanation:

For part a and b:

We have a system with two equations and two variables.

We have these data:

X = distance = 60m

t = time = 6.0s

Sf = Final speed = 15m/s

And We need to find:

So = Inicial speed

a = aceleration

We are going to use these equation:

$Sf^2=So^2+(2*a*x)$

$Sf=So+(a*t)$

We are going to put our data:

$(15m/s)^2=So^2+(2*a*60m)$

$15m/s=So+(a*6s)$

With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.

$\sqrt{(15m/s)^2-(2*a*60m)}=So$

$15m/s-(a*6s)=So$

$\sqrt{(15m/s)^2-(2*a*60m)}=15m/s-(a*6s)$

$[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}$

$(15m/s)^2-(2*a*60m)}=(15m/s)^{2}-2*(a*6s)*(15m/s)+(a*6s)^{2}$

$-120m*a=-180m*a+36s^{2}*a^{2}$

$0=120m*a-180m*a+36s^{2}*a^{2}$

$0=-60m*a+36s^{2}*a^{2}$

$0=(-60m+36s^{2}*a)*a$

$0=a1$

$\frac{60m}{36s^{2}} = a2$

$1.67m/s^{2}=a2$

If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2

After, we are going to determine the speed in the first point:

$Sf=So+(a*t)$

$15m/s=So+1.67m/s^2*6s$

$15m/s-(1.67m/s^2*6s)=So$

$4.98m/s=So$

For part c:

We are going to use:

$Sf^2=So^2+(2*a*x)$

$(4.98m/s)^2=0^2+(2*(1.67m/s^2)*x)$

$\frac{24.80m^2/s^2}{3.34m/s^2}=x$

$7.42m=x$

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2 years ago

A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg

marissa [1.9K]

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

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2 years ago

A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2

LiRa [457]

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

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2 years ago

A 4.99 m long rod of negligible weight is attached on one end to a ball joint which allows the rod to rotate in all directions.

Aleksandr [31]

Answer:

The answer is 91.18 Nm

Explanation:

Solution

Recall that

The length of the rod = 4.99 m

∅ = 26°

Force = 62.5N

Now,

T = r * F

The direction of the torque will be in horizontally northward

The torque magnitude is T =r F sin θ

where ∅ will be the angle between r + F θ= 163°

Therefore,

T = 4.99 * 62.5 * sin 163

T =91.18 Nm

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2 years ago