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iogann1982 [59]

2 years ago

10

All other things being equal, current will be larger in a wire that has a larger value of All other things being equal, current

will be larger in a wire that has a larger value of:______
a. conductivity.
b. resistivity.
c. the coefficient of current net charge.
d. potential.

1 answer:

morpeh [17]2 years ago

7 0

Answer:

a) conductivity

Explanation:

Assuming the wire obeys Ohm's Law, for a given voltage, in order to have a larger current, the resistance must be lower.
The resistance of a resistive element is given by the following equation:

$R = \rho * \frac{L}{A}$

If geometric parameters keep the same, the only choice to lower the resistance value is diminishing the resistivity ρ.
The resistivity, indeed, is defined as the inverse of the conductivity, σ.
So, if resistivity decreases, conductivity increases.
So, in order to have a larger current, all other things being equal, the wire must have a larger value of conductivity.
The option a) is the right one.

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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

$V=X/T$

First stage:

T1=5s

$v_{f} =v_{o} - at$

But, $v_{f} =0$ (decelerates to rest)

then: $a =v_{o} /t=0.3/5=0.06m/s^{2}$

on the other hand:

$x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m$

X1=75cm

Second stage:

T2=5s

$x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m$

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0

2 years ago

In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad

Crank

Answer:

a) Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c) Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

We replace

ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

Q = A v

Q = A₁ v₁

Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

A₁ = π r₁

A₂ = π r₂

We replace

Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

r₁ = d₁ / 2 = 1.00 / 2

r₁ = 0.500 10⁻² m

r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

Q = 1.57 10⁻² √(12 10³/2550)

Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

Q = 1.57 10⁻² √(2 12 10³ / (850 3)

Q = 4.8 10⁻² m³ / s

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2 years ago

If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?

natulia [17]

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An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

FrozenT [24]

Answer:

The total distance traveled is 28 inches.
The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector $\hat{i}$ pointing in the west direction, the ant start at position

$\vec{r}_0 = 16 \ inch \ \hat{i}$

Then, moves to

$\vec{r}_1 = 29 \ inch \ \hat{i}$

so, the distance traveled here is

$d_1 = |\vec{r}_1 - \vec{r}_0 | = | 29 \ inch \ \hat{i} - 16 \ inch \ \hat{i} |$

$d_1 = | 13 \ inch \ \hat{i} |$

$d_1 = 13 \ inch$

after this, the ant travels to

$\vec{r}_2 = 14 \ inch \ \hat{i}$

so, the distance traveled here is

$d_2 = |\vec{r}_2 - \vec{r}_1 | = | 14 \ inch \ \hat{i} - 29 \ inch \ \hat{i} |$

$d_2 = | - 15 \ inch \ \hat{i} |$

$d_2 = 15 \ inch$

The total distance traveled will be:

$d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch$

The displacement is the final position vector minus the initial position vector:

$\vec{D}=\vec{r}_2 - \vec{r}_1$

$\vec{D}= 14 \ inch \ \hat{i} - 16 \ inch \ \hat{i}$

$\vec{D}= - 2 \ inch \ \hat{i}$

This is 2 inches to the east.

6 0

2 years ago