A speedboat initially at rest accelerates at 4.0 m/s (squared) for 7.0 s. How far does the speedboat move in 7.0 s?

A ringing bell sends sound waves in all directions places sides

Strike441 [17]

Answer:

direction

Explanation:

because particles surround the bell, so when the bell vibrates, it causes particles surrounding it to vibrate back and forth vigorously. as these particles vibrate they collide with the neighbouring particles, passing on the energy.

hope this is what you are asking, if not please report it so that someone else gets to try it.

4 0

3 years ago

A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

Where:

N = is the number of turns = 1

ΔΦ = ΔB*A

Δt = is the time = 0.3 s

A = is the loop of wire area = πr² = πd²/4

ΔB: is the magnetic field = (0 - 1.04) T

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

3 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

What happens to the particles in water as the water is heated and turns to vapor? (2 points)

Naddik [55]

Answer:

The particles will more likely to move faster since they are converted from a liquid to gas.

Rules for States of Matter:

1. Solid particles always are packed close together and don't have much space to move.

2. Liquid particles have space to move around but are still packed together, but not as close as solid.

3. Gas particles are moving freely, in fact they are in the air! Gas particles are free to move wherever. For example, the air has gas particles that are constantly bumping into each other.

Let me know if I am right =)

4 0

3 years ago

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$