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3 years ago

13

What were the first batteries called?

2 answers:

MrRa [10]3 years ago

7 0

Correct answer is A
Voltaic Piles

Nitella [24]3 years ago

5 0

Answer:

A. Voltaic pile

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A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficien

Neporo4naja [7]

Answer:

The magnetic field is $B = 8.20 *10^{-3} \ T$

Explanation:

From the question we are told that

The mass of the metal rod is $m = 0.12 \ kg$

The current on the rod is $I = 4.1 \ A$

The distance of separation(equivalent to length of the rod ) is $L = 6.3 \ m$

The coefficient of kinetic friction is $\mu_k = 0.18$

The kinetic frictional force is $F_k = 0.212 \ N$

The constant speed is $v = 5.1 \ m/s$

Generally the magnetic force on the rod is mathematically represented as

$F = B * I * L$

For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so

$F_ k = B* I * L$

=> $B = \frac{F_k}{L * I }$

=> $B = \frac{0.212}{ 6.3 * 4.1 }$

=> $B = 8.20 *10^{-3} \ T$

7 0

3 years ago

A solid weighs 200N in air, 150N in water and 170N in a liquid. Find relative density of solid, relative density of liquid and d

fomenos

Answer:

$\rho_{s} = 4$

$\rho_{l} = 0.6$

$\rho{liq} = 600 kg/m^{3}$

Given:

Weight of solid in air, $w_{sa} = 200 N$

Weight of solid in water, $w_{sw} = 150 N$

Weight of solid in liquid, $w_{sl} = 170 N$

Solution:

Calculation of:

1. Relative density of solid, $\rho_{s}$

$\rho_{s} = \frac{w_{sa}}{w_{sa} - w_{sw}}$

$\rho_{s} = \frac{200}{200 - 150} = 4$

2. Relative density of liquid, $\rho_{l}$

$\rho_{l} = \frac{w_{sa} - w_{sl}}{w_{sa} - w_{sw}}$

$\rho_{l} = \frac{200 - 170}{200 - 150} = 0.6$

3. Density of liquid in S.I units:

Also, we know:

$\rho{l} = \frac{\rho_{liq}}{\rho_{w}}$

where

$= {\rho_{liq}}$ = density of liquid

$= {\rho_{w}} = 1000 kg/m^{3}$ = density of water

Now, from the above formula:

$0.6 = \frac{\rho_{liq}}{1000}$

$\rho{liq} = 600 kg/m^{3}$

3 0

3 years ago

How can you define a solution to an equation?

sleet_krkn [62]

A solution is a value or a collection of values.. when substituted for the unknowns, the equation become an equality.
Example : x + 2 = 7
When we out the 5 in place of x we get: 5 + 2 = 2

8 0

3 years ago

Which of the following is an x-intercept of the function below

Assoli18 [71]

Answer:

Intercepts occur at x = 1, y = 0

x = 3, y = 0

Only (D) is correct

4 0

3 years ago

An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust

zubka84 [21]

Answer:

Time to Reach Saturation = 0.0146 day

Explanation:

In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:

Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³

Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³

Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³

Now, we find volume flow rate of air through filter:

Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³

Volume Flow Rate of Air = 11.33 m³/min

Now, we calculate rate of dust filtered:

Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)

Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)

Rate of Dust Filtered = 2.38 mg/min

Now, for the time required to reach saturation:

Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)

Time to Reach Saturation = (50 mg)/(2.38 mg/min)

Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)

<u>Time to Reach Saturation = 0.0146 day</u>

3 0

3 years ago