P.E QUESTION

Explain whether a particle moving in a straight line with constant speed does or does not have an acceleration. b) Explain wheth

Lelechka [254]

Answer:

A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

Please Give Brainliest

5 0

2 years ago

What percentage of all traffic deaths occur in cars traveling less than 60 mph?

Alina [70]

This is a bit too broad. Maybe this is something to google up

5 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

When you strike two tuning forks simultaneously, you hear three beats per second. The frequency of the first tuning fork is 440

tekilochka [14]

Answer:4

Explanation:

Given

Frequency of beats is 3 beats per second

Frequency of first tuning fork is $f_1=440\ Hz$

Beat frequency is difference in the frequency of tuning forks i.e.

either $f_1-f_2$ or $f_2-f_1 =3$

so $f_2=3+440=443\ Hz$

or

$f_2=440-3=437\ Hz$

so there is not enough information given to decide.

5 0

3 years ago

Read 2 more answers

Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?

ozzi

Answer: The forces acting on both of them will increase in magnitude.

Explanation:

According to Coulomb's law, the electrostatic force between two bodies is proportional to the product of their two charges. If the charge on A is increased this product increases in size (it must have been non-zero to begin with, since there was a force between them at first). Thus, the force between them rises.

6 0

3 years ago