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3 years ago

When must batteries in an emergency locator transmitter (ELT) be replaced or recharged, if re-chargeable?

Physics

1 answer:

tatiyna3 years ago

6 0

Answer:

After one half of the battery's useful life.

Explanation:

Batteries of the emergency locator transmitter (ELT) must be replaced or recharged after one half of the battery's useful life because if it is exposed to the high temperature for a long period of time such as the air plane parked in the sun will result in the deterioration of battery which may makes the transmitter out of order before the expiry date of the battery. So it will be safe to do that after the use of one half of the battery's life.

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Two protons, starting several meters apart, are aimed directly at each other with speeds of 2.00 * 105 m>s, measured relative

victus00 [196]

Explanation:

Since, the two protons are at the initial state of point a and point b. Hence, total mechanical energy at point a and point b is as follows.

$U_{a} + K_{a} = U_{b} + K_{b}$

or, $K_{a} = U_{b}$

where, $K_{a}$ = kinetic energy of two protons

So, $2(\frac{1}{2}mV^{2}_{a}) = \frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r_{b}}$

$r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

Putting the given values into the above formula we will calculate the value of $r_{b}$ as follows.

$r_{b} = \frac{1}{4 \pi \epsilon_{o}} \frac{e^{2}}{mV^{2}_{a}}$

= $(9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}^{2} C}{1.67 \times 10^{-27} kg \times 2 \times 10^{5}}$

= $3.45 \times 10^{-12} m$

Now, we will calculate the maximum electric field as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{|e|^{2}}{r^{2}_{b}}$

= $9.0 \times 10^{9} Nm^{2}/C^{2} \frac{(1.602 \times 10^{-19}C)^{2}}{3.45 \times 10^{-12}}$

= $0.194 \times 10^{-4} N$

Therefore, we can conclude that the maximum electric force that these protons will exert on each other is $0.194 \times 10^{-4} N$ .

7 0

3 years ago

Three equal 1.60-μCμC point charges are placed at the corners of an equilateral triangle with sides 0.800 mm long. What is the p

Pepsi [2]

Answer:

= 0.086J

Explanation:

we are given three equal charges q₁ q₂ q₃ = 1.60 μC

sides of triangle = 0.800m

U = 1/4πε₀(3q²/r)

1/4πε₀ = 9.0 × 10⁹ N.m²/C²

U = 9.0 × 10⁹ ((3 * 1.60 × 10⁻⁶)² / (0.800))

= 0.086J

3 0

3 years ago

नारायणको तौल 70kg छ । उसले 3m उचाइ भएको रूख चढ्न 20 sec मा लाग्छ भने

Serga [27]

Sorry I tbh don’t understand wat ur saying

7 0

3 years ago

An object's movement around an internal axis is _____.

zaharov [31]

Answer:

brooo it is rotation .............

4 0

3 years ago

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0

3 years ago