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3 years ago

What is the total amount of work required to

Physics

1 answer:

ASHA 777 [7]3 years ago

7 0

Answer:

$W = 1.6 \times 10^{-17} J$

Explanation:

Work done to move a charge through a given potential difference is

$W = q(\Delta V)$

here

q = magnitude of charge

$\Delta V$ = potential difference

here it is given that charge of the particle is same as charge of proton

$q = 1.6 \times 10^{-19} C$

$\Delta V = 100 V$

now we have

$W = (1.6 \times 10^{-19}) (100)$

$W = 1.6 \times 10^{-17} J$

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Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has

Degger [83]

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

6 0

3 years ago

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

stiv31 [10]

Answer:

Magnetic field, $B=2\times 10^{-4}\ T$

Explanation:

Given that,

Velocity of electron, $v=5\times 10^7\ m/s$

It enters a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, $E=10^4\ V/m$

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, $F_m=qvB\ sin\theta$ .......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

$qvB\ sin\theta=qE$

$B=\dfrac{E}{v}$

$B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}$

B = 0.0002 T

$B=2\times 10^{-4}\ T$

Hence, this is the required solution.

3 0

3 years ago

How many coulombs of positive charge are there in 47.0 gm of plutonium, given its atomic mass is 244 and that each plutonium ato

enot [183]

Answer:

1.78×10⁶ C

Explanation:

Using the atomic mass of pluonium atoms (244 g/mol), you can calculate the number of atoms in 47.0 g. Then, knowing that each plutonium atom has 96 protons, you calculate the number of protons in the 47.0 g sample. Finally, using the positive charge of one proton, you calculate the total positive charge in the 47.0 g of plutonium.

1. Number of atoms of plutonium in 47.0 g

Number of moles = mass / atomic mass = 47.0 g / 244 = 0.1926 moles

Number of atoms = number of moles × 6.022 × 10²³ atoms/mol

Number of atoms = 0.1926 mol × 6.022 × 10²³ atoms/mol = 1.15998×10²³ atoms

2. Number of protons

Number of protons = 1.15998×10²³ atoms × 96 protons/atom = 1.11385×10²⁵ protons

3. Charge

Charge = charge of one proton × number of protons

Charge = 1.602×10⁻¹⁹ C/proton × 1.11385×10²⁵ protons = 1.78×10⁶C

7 0

3 years ago

I need help with part D

natka813 [3]

1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.

8 0

3 years ago

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

hammer [34]

Answer:

1201 lbs

Explanation:

Given that in mammals, the weight of the heart is approximately 0.5% of the total body weight.

Let the weight of the heart of a mammal be H

And the weight of the total body be B

The linear model that can gives the heart weight in terms of the total body weight will be:

H = 0.005B

B.) To find the weight of the heart of a whale whose weight is 2.402 × 105 lbs, substitute the whole weight in the formula.

H = 0.005 × 2.402 × 10^5

H = 1201 lbs

Therefore, the weight of the heart of the whale is 1201 lbs

8 0

3 years ago