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Morgarella [4.7K]
3 years ago
5

Which type of hearing problem can be reduced which ordinary hearing aids

Physics
1 answer:
Paul [167]3 years ago
8 0
Conduction deafness.
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When two license plates are issued they need to be attached to the _________ of the vehicle they were issued for.
Sergio039 [100]

Answer:

Front and the rear bumpers.

Explanation:

When two license plates are issued they need to be attached to the Front and the rear bumpers of the vehicle they were issued for.

7 0
3 years ago
Question 8
Marrrta [24]

Answer:

It has a mass of 40 kg.

Explanation:

Because Force = mass x Acceleration or F = m a, we could say that the mass is force/acceleration which in your case is 2,400/60 which equals 40 kg.

3 0
2 years ago
If a wave hits a smooth surface at an angle of incidence of 40 degrees, the angle of reflection is
dem82 [27]
As the surface given is a smooth surface, we can use specular reflection. According to the law of specular reflection, the angle of incidence equals the angle of reflection, so it will also be 40°. Answer is A.
5 0
3 years ago
Read 2 more answers
What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffracti
Artist 52 [7]

Answer:

The wavelength is 3500 nm.

Explanation:

d= \frac{1}{700 lines per mm} = 0.007mm = 7000 nm

n= 1

θ= 30°

λ= unknown

Solution:

d sinθ = nλ

λ = \frac{7000 nm sin 30}{1}

λ = 3500 nm

8 0
3 years ago
A tank having a volume of 0.85 m 3 initially contains water as a two-phase liquid-vapor mixture at 260 o C and a quality of 0.7.
Keith_Richards [23]

Answer:=14,160 kJ

Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below

Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase

C MPa m^3/kg kJ/kg kJ/kg

1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture

2 260 4.689 0.0422 2599 2797 1 Saturated Vapor

The mass initially contained in the tank is m1 = V/v1

m1 =0.85 m^3 /0.02993 m^3 /kg

= 28.4 kg

The mass finally contained in the tank is

m2 =V2/v

= 0.85 m^3 /0.0422 m^3 /kg

= 20.14 kg

The heat transfer is then

Qcv = m2u2 − m1u1 − he(m2 − m1)

Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ

4 0
3 years ago
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