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2 years ago

A stock person at the local grocery store has a job consisting of the following five segments:

Physics

1 answer:

vaieri [72.5K]2 years ago

6 0

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

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1 year ago

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Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with

mr_godi [17]

A challenge scientists face with this process is the use of ultrathin iron oxide, to pull protons off water and produce hydrogen gas, which itself is a poor electrical conductor.

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1 year ago

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Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac

dangina [55]

Explanation:

Relation between electric field and charge density is as follows.

E = $\frac{\sigma}{2 \epsilon}$

where, $\sigma$ = charge density

$\epsilon$ = permittivity of free space = $8.85 \times 10^{-12}$

So, $E_{\text{outside}}$ = 0

$E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}$

or, $E_{inside} = \frac{\sigma}{\epsilon}$

Now, formula to calculate the potential difference of two conductors is as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

It is given that,

d = 6.0 mm = $6 \times 10^{-3} m$

$\sigma = 40 \times 10^{-12} C/m^{2}$

Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

= $\frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}$

= 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

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2 years ago

Burning coal, which is how many power plants generate electricity, releases a number of harmful byproducts. Particulate pollutio

kiruha [24]

Answer:

$7.25\times 10^{-16}\ C$

Explanation:

F = Force = $7.25\times 10^{-11}\ N$

E = Electric field = $1\times 10^5\ N/C$

Force is given by

$F=qE$

$\Rightarrow q=\dfrac{F}{E}$

$\Rightarrow q=\dfrac{7.25\times 10^{-11}}{1\times 10^5}$

$\Rightarrow q=7.25\times 10^{-16}\ C$

The charge on the particle is $7.25\times 10^{-16}\ C$

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2 years ago

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they? (

valkas [14]

The gravitational force between two masses m₁ and m₂ is
$F=G \frac{m_{1} m_{2}}{d^{2}}$
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm

8 0

2 years ago