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2 years ago

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

Physics

1 answer:

tatiyna2 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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A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of gla

Diano4ka-milaya [45]

Complete Question

A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,

A :

a force repels the glass out of the capacitor.

B :

a force attracts the glass into the capacitor.

C :

no force acts on the glass.

D :

a net charge appears on the glass.

E :

the glass makes the plates repel each other.

Answer:

The correct option is B

Explanation:

Generally when the glass dielectric is slowly inserted between the plated,

The positive plate of the capacitor will induce a negative charge on the glass while the negative plate of the capacitor will induce a positive charge on glass which a electric field that posses an electric force that will attract the glass

3 0

3 years ago

If the given wave has a frequency of 100 Hz, what frequency will the sixth harmonic have?

alukav5142 [94]

Answer:

600Hz

Explanation:

In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.

In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz

4 0

3 years ago

Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.

Anika [276]

Answer : The significant digit is 6

Explanation :

Multiply $5.036\times10^{2}m$ by $0.078\times10^{-1}m$

Now, on multiplying

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}$

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}$

Now, the significant digit is 6.

Hence, this is the required solution.

3 0

2 years ago

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo

NISA [10]

Explanation:

The given data is as follows.

Mass of small bucket (m) = 4 kg

Mass of big bucket (M) = 12 kg

Initial velocity ( $v_{o}$ ) = 0 m/s

Final velocity ( $v_{f}$ ) = ?

Height $H_{o} = h_{f}$ = 2 m

and, $H_{f} = h_{o}$ = 0 m

Now, according to the law of conservation of energy

starting conditions = final conditions

$\frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}$

$\frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)$

235.44 = $8V^{2}_{f}$ + 78.48

$V_{f}$ = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0

2 years ago

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of

Elan Coil [88]

Given :

A spring with a spring constant of 1730 N/m is compressed 0.136 m from its equilibrium position with an object having a mass of 1.72 kg.

To Find :

The embankment in the height.

Solution :

Since no external force is acting in the system, therefore total energy will be conserved.

Initial kinetic energy of the object = Energy stored in spring

$K.E _i = \dfrac{kx^2}{2}\\\\K.E_i = \dfrac{1730\times 0.136^2}{2}\\\\K.E_i = 16\ J$

Also, initial potential energy is 0.

Now,

$K.E_i + P.E_i = K.E_f + P.E_f\\\\16 + 0 = \dfrac{1.72\times 2.45^2}{2}+ mgh\\\\mgh =16 - 5.16\\\\h = \dfrac{16 - 5.16}{1.72 \times 9.8}\\\\h = 0.64\ m$

Therefore, the embankment height is 0.64 m.

3 0

3 years ago