Which statement accurately describes static electricity?

Find the interest. A 105-day note for $14,200 at 8.5% interest.

mr Goodwill [35]

<h3><u>Answer</u>;</h3>

$347.22

<h3><u>Explanation</u>;</h3>

Principal = $14,200

Rate = 8.5%

Time = 105 days = 105/365

Interest = Principal x Rate x Time

Interest = 14,200 x 0.085 x 105/365

Interest = 347.219

= $347.22

7 0

3 years ago

With a diameter that's 11 times larger than Earth's, _______ is the largest planet.

katovenus [111]

With a diameter that's 11 times larger than Earth's, Jupiter is the largest planet

6 0

3 years ago

An athlete in a gym applies a constant force of 50 N to the pedals of a bicycle to keep the rotation rate of the wheel at 10 rev

marishachu [46]

Answer:

Explanation:

Given that,

Force applied to pedal F = 50N

Angular velocity ω = 10rev/s

We know that, 1rev = 2πrad

Then, ω = 10rev/s = 10×2π rad/s

ω = 20π rad/s

Length of pedal r = 30cm = 0.3m

Power?

Power is given as

P = τ×ω

We need to find the torque τ

τ = r × F

Since r is perpendicular to F

Then, τ = 0.3 × 50

τ = 15 Nm

Then,

P = τ×ω

P = 15 × 20π

P = 942.48 Watts

power delivered to the bicycle by the athlete is 942.48 W

6 0

3 years ago

When you observe a distant galaxy whose photons have traveled for 10 billion years before reaching earth we are seeing that gal?

expeople1 [14]

In reality we don't see the galaxy we see it's reflection .. the light hits or got emitted by the star travel all the way long to hit our eyes .. we see their reflection . everything around you that you see is it's reflection

8 0

3 years ago

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

3 years ago