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3 years ago

15

If there are 60 min in 1hr, and in 1min there are 60 sec; Solve the following problem using dimensional analysis.: 3,800 hrs. to

seconds(sec)

1 answer:

GREYUIT [131]3 years ago

6 0

13680000 should be the number of seconds

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I need help with 23 please.

Ne4ueva [31]

The answer is c.

hope this helps! :)

8 0

3 years ago

A car is traveling north at 17.7m/s After 12 s its velocity is 14.1m/s in the same direction. Find the magnitude and direction o

Nat2105 [25]

The car is initially traveling north at 17.7 m/s, and after 12 s, its velocity is 14.1 m/s, still due north. This means that the direction of the car has not changed, so we can already say that the direction of the acceleration is north (if the magnitude of the acceleration is positive) or south (if the magnitude of the acceleration is negative).

To find the magnitude of the average acceleration, we must calculate the ratio between the change in velocity and the time taken:

$a=\frac{v_f -v_i}{t}=\frac{14.1 m/s-17.7 m/s}{12 s}=-0.3 m/s^2$

Since the acceleration is negative, it means it is in the opposite direction to the motion of the car, therefore south. Therefore, the correct answer is

b) 0.30 m/s2, south

7 0

3 years ago

The specific gravity of a product can be found on the material safety data sheet. how do you know if a product will sink in wate

Kryger [21]

Specific gravity is the ratio between the substance's density and the density of water. Being a ratio of dimensions with same units of measure, this value is unitless.

The substance or objects sinks in water if its density is greater than that of water. On the other hand, it floats if its density is lesser compared to that of water.

Having said so, the substance or object sinks in water if the specific gravity is greater than 1 and floats if the specific gravity is lesser than 1.

sinks: SG > 1
floats: SG < 1

8 0

3 years ago

2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust

notsponge [240]

Answer:

Explanation:

This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

7 0

3 years ago

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

3 years ago