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3 years ago

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Calculate the electric field at the location of a charge if the electric force is 10 newtons and the charge on the particle is 6

coulombs.
a. 0.6 j
b. 0.6 n/c
c. 1.7 v
d. 1.7 n/c

1 answer:

kolezko [41]3 years ago

6 0

I think it may be c, 1.7 v. I work with electricity a lot.

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2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo

seropon [69]

Answer:

The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

$\omega=376.9\ rad/s$

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

$B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}$

$B_{max}=2.901\times10^{-13}\ T$

Hence, The maximum value of the induced magnetic field is $2.901\times10^{-13}\ T$ .

7 0

3 years ago

A constant power is supplied to a rotating disc .the relationship of angular velocity of disc and number of rotations made by th

Bogdan [553]

A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion.

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3 years ago

Complete the equation???

soldi70 [24.7K]

Here in nuclear reaction we can say that sum of neutrons and protons in reactant side and product side will be same always

Here mass number on the product side is given to us

so sum of mass number is given as

$A_1 + A_2 = 265 + 1 = 266$

now on the reactant side also the number must be same

$A_1' + A_2' = 58 + x$

now we will have

$58 + x = 266$

$x = 208$

Now number of protons on product side is given as

$P_1 + P_2 = 108 + 0$

now we also know that atomic number of Fe is 26

so now we will have

$P_1' + P_2' = 108$

$26 + P_2' = 108$

$P_2' = 82$

now the equation is given as

$_{26}^{58}Fe + _{82}^{208}Pb = _{108}^{266}Hs + _0^1X$

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2 years ago

Should i beat up my sibling cause they mean to me?

Vlada [557]

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3 years ago

A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc

kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

$\huge\boxed{\Sigma \tau = I \alpha}$

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

$\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}$

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2 years ago