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3 years ago

A charged ball of mass 10g and charge 2 C is placed at rest in a field of field strength 5NC as shown.

Physics

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:

1000 m/s

Explanation:

Given that

Mass of the ball, m = 10 g = 0.01 kg

Charge of the ball, q = 2 C

Field strength of the field, E = 5 N/C

Acceleration of the ball, a = ? m/s²

We know the the equation holds,

F = qE, on substituting, we have

F = 2 C * 5 N/C

F = 10 N

Also, we know that

F = ma, and as such,

a = F / m, on substituting we have

a = 10 N / 0.01 kg

a = 1000 m/s²

Therefore, the acceleration of the ball is 1000 m/s²

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3 years ago

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If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti

Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

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$\frac{T^2}{a^3}=const.$

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a is the semi-major axis of the orbit

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Therefore, the period of the orbit would increase.

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3 years ago

A copper wire has a square cross section 2.1 mm on a side. The wire is 4.3 m long and carries a current of 3.6 A. The density of

aleksley [76]

Explanation:

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3 years ago

While on a playground, you and your niece take turns sliding down a frictionless slide. Your mass is 75 kg while your little nie

Olenka [21]

Answer:

Explanation:

In absence of friction, total mechanical energy must be conserved.
This means that the change in gravitational potential energy (in magnitude) must be equal to the change in kinetic energy:

$\Delta K = \Delta U \\ \\ \frac{1}{2} * m* v^{2} = m*g*h$

As it can be seen, the mass m is on the both sides of the equation, which means that it can be simplified.
We can solve this equation for v (speed at the bottom) as follows:

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3 years ago

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

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3 years ago