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3 years ago

5

When electromagnetic fields interact with charged particles

1 answer:

dedylja [7]3 years ago

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A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

ElenaW [278]

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency c/λ where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

7 0

3 years ago

A block weighing 3.7 kg is suspended from the ceiling of a truck trailer by a hanging bungee cord. The cord has a cross-sectiona

sweet [91]

Answer:

$Y = 2.27 \times 10^{10} N/m^2$

Explanation:

Natural length of the string is given as

$L_o = 43 cm$

length of the string while block is hanging on it

$L = 53 cm$

extension in length is given as

$\Delta L = 10 cm$

now we have strain in the string is given as

$strain = \frac{\Delta L}{L}$

$strain = \frac{{10 cm}{43 cm}$

$strain = 0.23$

similarly we will have cross-sectional area of the string is given as

$A = 40 \times 10^{-6} m^2$

now the stress in the string is given as

$Stress = \frac{T}{A}$

$Stress = \frac{mg}{A}$

$Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}$

$stress = 9.07 \times 10^5 N/m^2$

Now Young's Modulus is given as

$Y = \frac{stress}{strain}$

$Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}$

$Y = 2.27 \times 10^{10} N/m^2$

5 0

3 years ago

A motorcycle and rider have a total mass equal to 300 kg. The rider applies the brakes, causing the motorcycle to decelerate at

seraphim [82]

Answer:

Net force = - 1500 N

Explanation:

We calculate the net force acting using Newton's second Law:

$F_{net}=m*a\\F_{net}=(300 \,kg)*(-5\,m/s^2)\\F_{net}=-1500\,N$

5 0

3 years ago

Direct current made it possible to distribute electric power over greater areas.

Mademuasel [1]

True, I'm not the best when it comes to science, but I'm pretty sure it's this

3 0

3 years ago

A 1,550 kg car moving south at 10.0 m/s collides with a 2,550 kg car moving north. The cars stick together and move as a unit af

nikitadnepr [17]

Answer:

14.47 m/s

Explanation:

The momentum must be preserved before and after the collision:

The total momentum before the collision

$m_1v_1 + m_2v_2$

where $m_1 = 1550 kg, m_2 = 2550 kg$ are the masses of the car moving south and north, respectively, before the collision. $v_1 = -10m/s$ is the velocity of the car moving South. We take the velocity to the North as the positive direction

The total momentum after the collision

$(m_1 + m_2)V$

where V = 5.22m/s is the velocity of both cars after the collision

We can equalize the 2 equations and plug in the numbers:

$m_1v_1 + m_2v_2 = (m_1 + m_2)V$

$1550(-10) + 2550*v_2 = (1550 + 2550)5.22$

$-15500 + 2550v_2 = 21402$

$2550v_2 = 36902$

$v_2 = 36902/2550 = 14.47 m/s$

6 0

3 years ago