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ELEN [110]
3 years ago
14

An accelerometer is a device that uses the extension of a spring to measure acceleration in terms of Earth's gravitational accel

eration (g). What is the approximate acceleration if this accelerometer spring is extended just 0.43 centimeters?
Physics
1 answer:
Aloiza [94]3 years ago
8 0

letter B for plato is correct!!

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As you stand near a railroad track, a train passes by at a speed of 31.7 m/s while sounding its horn at a frequency of 218 Hz. W
Darya [45]

Explanation:

Given that,

Frequency of train horn, f = 218 Hz

Speed of train, v_t = 31.7 m/s

The speed of sound, V = 344 m/s (say)

The speed of the observed person, V_o=0\ m/s

(a) When the train approaches you, the Doppler's effect gives the frequency as follows :

f'=f(\dfrac{V}{V-v_t})\\\\f'=218\times (\dfrac{344}{344-31.7})\\\\f'=240.12\ Hz

(b) When the train moves away from you, the Doppler's effect gives the frequency as follows :

f'=f(\dfrac{V}{V+v_t})\\\\f'=218\times (\dfrac{344}{344+31.7})\\\\f'=199.6\ Hz

Hence, this is the required solution.

6 0
4 years ago
How much of the matter in the universe is comprised of atoms?
Deffense [45]

Explanation:

Atoms are the components of ordinary matter, also called baryonic matter, which only represents 4% of the universe, while the remaining 96% would be formed by what is known as dark matter and dark energy which constitute two of the unsolved problems in physics.

4 0
3 years ago
Read 2 more answers
A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equili
pentagon [3]

Answer:

x = 1.26 sin 3.16 t

Explanation:

Assume that the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

V=\dfrac{dx}{dt}

V= A ω cosωt

Maximum velocity

V(max)= Aω

Given that F= 32 N

F = K Δ

K=Spring constant

Δ = 0.4 m

32 =0.4 K

K = 80 N/m

We know that  ω²m = K

8 ω² = 80

ω = 3.16 s⁻¹

Given that V(max)= Aω = 4 m/s

3.16 A = 4

A= 1.26 m

Therefore the general equation of displacement

x = 1.26 sin 3.16 t

7 0
3 years ago
Read 2 more answers
Which property of sound waves decreases as the square of the distance from the source increases?
Vitek1552 [10]

Answer:

Intensity

Explanation:

The intensity of a sound wave is equal to the ratio between to the power emitted by the source divided by the area of the spherical surface through which the wave propagates:

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

where

P is the power

4\pi r^2 is the area of the spherical surface

r is the distance from the source

As we see from the formula, the intensity is inversely proportional to the square of the distance from the source:

I\propto \frac{1}{r^2}

so, intensity is the correct answer.

3 0
3 years ago
Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f
gulaghasi [49]

Answer:

A)3.8196 * 10^{9} Newtons        B) 2.153 * 10^{9} Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as  120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = πR^{2}.

Area Expansivity β =    \frac{Change in Area}{Original Area * Temperature Rise}

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = \frac{0.2m}{2} = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = \frac{0.15}{2} = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * 0.1^{2} =  0.0314m^{2}

Area of Rod BC = π * 0.075^{2} = 0.0177m^{2}.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9*10^{-6} * 0.0314 * 30 = 3.183 * 10^{-5}m^{2}.

For Rod BC we have = 2 * 16.9*10^{-6} * 0.0177 * 30 = 1.794∈-5m^{2}.

The forces on each rods will be given by.

Force AB = Pressure * Area  of Rod AB

                 = 120GPa * 3.183 * 10^{-5} gives 3.8196 * 10^{9} Newtons.

Force BC = Pressure * Area of Rod BC

                = 120GPa * 1.794 * 10^{-5} gives 2.153 * 10^{9} Newtons

7 0
3 years ago
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