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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi

jeka94

Answer:

I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

d = (d₁ + d₂) / 2

d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

M = M_man - 0.2 M_man = 80 (1-0.2)

M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis: Ic = ½ M r²

A rod that rotates at the end: I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

I₁ = 8.258 + 5.33 + 4.129

I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

I₂ = ½ M r² + 2 m D²

I₂ = ½ 64 0.508² + 2 8 0.508²

I₂ = 8,258 + 4,129

I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

I₁ / I₂ = 17,717 /12,387

I₁ / I₂ = 1.43

3 0

3 years ago

If you install special sound-reflecting windows that reduce the sound intensity level by 30.0 db , by what factor have you reduc

shusha [124]

Answer: The sound intensity was reduced by -14.77 dB.

Explanation:

The level of the intensity of the sound with respect to the reference value is called Sound Intensity level.

$L_I=10\log\frac{I}{I_o} dB$

I = Intensity of the sound

$I_o$ = Reference sound intensity

According to question, sound-reflecting windows that reduce the sound intensity level by 30.0 db .

Let the intensity of the sound be $I_o$ and reduced intensity be I.

$I=\frac{I_o}{30} dB$

$L_I=10\log\frac{I}{I_o} dB$

$L_I=10\log\frac{\frac{I_o}{30}}{I_o}=10\log\frac{1}{30} dB=-14.77 dB$

The sound intensity was reduced by -14.77 dB.

3 0

3 years ago

An office worker has his lunch on the 53rd floor of an office building. In which case is the gravitational potential energy of t

alekssr [168]

Answer: In all cases, gravitational potential energy, respect from a common reference level (street level for instance) is the same.

Explanation:

Gravitational potential energy, comes from the action of the gravitational field, which is a conservative one.

This means, that if someone does work against the field (for instance lifting a weight from the ground to a given height), and then drops the weight, the work done by the field will be exactly the same.

Otherwise, the object would have remaining energy after returning to the starting point, which would violate the laws of physics, specially Thermodynamic's 2nd Law.

For this reason, the gravitational potential energy only depends from the height, and is independent from the path taken to reach there.

0 0

4 years ago

What is the efficiency of an engine that exhausts 440 J of heat to a cold reservoir and receives 570 J of heat from a hot reserv

katrin2010 [14]

Answer:

Efficiency = 77%

Explanation:

Input energy = 570 J

Output energy = 440 J

To find the efficiency;

$Efficiency = \frac {Out-put \; energy}{In-put \; energy} * 100$

Substituting into the equation, we have;

$Efficiency = \frac {440}{570} * 100$

$Efficiency = 0.7719 * 100$

Efficiency = 77.19 ≈ 77%

Therefore, the efficiency of the engine is 77 percent.

7 0

3 years ago