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nasty-shy [4]
2 years ago
7

The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them

?
Physics
1 answer:
hammer [34]2 years ago
4 0

It is not only the horizontal part of the loop that dips into the magnetic field. We can neglect or disregard the horizontal parts of the loop that hollows into the magnetic fields since only the parts perpendicular to the magnetic field complement to it.

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What units are given to the right of the equals sign
zhuklara [117]
The answer

2y + 14 = 17

The 17 is to the right of the = sign
It is also the answer
7 0
2 years ago
A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
2 years ago
A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th
mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

6 0
2 years ago
Calculate the pressure, in atmospheres, exerted by each of the following:
gregori [183]

Answer:

a) 14.2 atm

b) 4.46 atm

c) 1.06 atm

Explanation:

For an ideal gas,

PV = nRT

P = pressure of the gas

V = volume occupied by the gas

n = number of moles of the gas

R = molar gas constant = 0.08206 L.atm/mol.K

T = temperature of the gas in Kelvin

a) For HF,

P =?, V = 2.5L, n = 1.35 moles, T = 320K

P = 1.35 × 0.08206 × 320/2.5

P = 14.2 atm

b) For NO₂

P =?, V = 4.75L, n = 0.86 moles, T = 300K

P = 0.86 × 0.08206 × 300/4.75

P = 4.46 atm

c) For CO₂

P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K

P = 2.15 × 0.08206 × 330/55

P = 1.06 atm

4 0
2 years ago
A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla
ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

4 0
2 years ago
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