Just need help with this one

The strength of an electric field depends on the

KengaRu [80]

<span>The correct option is D) both A and B. In fact, the strength of an electric field produced by a single point charge is given by
</span> $E=k \frac{Q}{r^2}$ <span>
where Q is the charge that generates the field, k is the Coulomb's constante and r is the distance from the charge source of the field.

From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>

5 0

3 years ago

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Determine what forces are present in this situation: An elevator is rising at a constant speed; the elevator is the object in th

Alecsey [184]

Tension pointing up at the top Weight pointing down at the bottom If there is someone standing inside the elevator, there will be a normal reaction force pointing down at the bottom.

7 0

3 years ago

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A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in

Leya [2.2K]

Answer:

The speed of its center of mass = $\sqrt{\frac{3}{2}gL}$

Explanation:

Consider the potential energy at the level of center of mass of rod below the pivot=0

Mass of uniform rod=M

Length of rod=L

The rotational inertia about the end of a uniform rod= $\frac{1}{3}ML^2$

Kinetic energy at the level of center of mass of rod below the pivot= $\frac{1}{2}I\omega^2$

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

$mgh+0=0+\frac{1}{2} I\omega^2$

Where $K.E=\frac{1}{2} I\omega^2$

I=Moment of inertia

$\omega$ =Angular velocity

Substitute the values then we get

$MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2$

$\omega^2=\frac{6g}{L}$

Now, we know that $\omega=\frac{v}{r}$ , $r=\frac{L}{2}$

Substitute the values then we get

$\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}$

$\frac{v^24}{L^2}=\frac{6g}{L}$

$v^2=\frac{6g\times L^2}{4L}$

$v^2=\frac{3gL}{2}$

$v=\sqrt{\frac{3}{2}gL}$

Hence, the speed of its center of mass = $\sqrt{\frac{3}{2}gL}$

4 0

4 years ago

A(n) ____ like rubber or plastic has high electrical resistance.

blagie [28]

Answer:

insulator

Explanation:

6 0

2 years ago

Two students hear the same sound and their eardrums receive the same power from the sound wave. The sound intensity at the eardr

Margaret [11]

Answer:

a. d₁/d₂ = 1.09 b. 0.054 mW

Explanation:

a. What is the ratio of the diameter of the first student's eardrum to that of the second student?

We know since the power is the same for both students, intensity I ∝ I/A where A = surface area of ear drum. If we assume it to be circular, A = πd²/4 where r = radius. So, A ∝ d²

So, I ∝ I/d²

I₁/I₂ = d₂²/d₁² where I₁ = intensity at eardrum of first student, d₁ = diameter of first student's eardrum, I₂ = intensity at eardrum of second student, d₂ = diameter of second student's eardrum.

Given that I₂ = 1.18I₁

I₂/I₁ = 1.18

Since I₁/I₂ = d₂²/d₁²

√(I₁/I₂) = d₂/d₁

d₁/d₂ = √(I₂/I₁)

d₁/d₂ = √1.18

d₁/d₂ = 1.09

So, the ratio of the diameter of the first student's eardrum to that of the second student is 1.09

b. If the diameter of the second student's eardrum is 1.01 cm. how much acoustic power, in microwatts, is striking each of his (and the other student's) eardrums?

We know intensity, I = P/A where P = acoustic power and A = area = πd²/4

Now, P = IA

= I₂A₂

= I₂πd₂²/4

= 1.18I₁πd₂²/4

Given that I₁ = 0.58 W/m² and d₂ = 1.01 cm = 1.01 × 10⁻² m

So, P = 1.18I₁πd₂²/4

= 1.18 × 0.58 W/m² × π × (1.01 × 10⁻² m)²/4

= 0.691244π × 10⁻⁴ W/4 =

2.172 × 10⁻⁴ W/4

= 0.543 × 10⁻⁴ W

= 0.0543 × 10⁻³ W

= 0.0543 mW

≅ 0.054 mW

3 0

3 years ago