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3 years ago

15

A vehicle has an initial velocity of 30 meters per second. 30 seconds later, it is traveling at a velocity of 60 meters per seco

nd. What is the acceleration of the vehicle?

1 answer:

Over [174]3 years ago

3 0

1 m/s^2 (60-30) then divide by 30 secs :)

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Which of the following is not true?

ahrayia [7]

The answer is either C or D..

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3 years ago

A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.

alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Here,

k = Spring constant

m = Mass

Our values are given as,

$m = 35g = 35*10^{-3}kg$

$f = 1 Hz$

Rearranging to find the spring constant we have that,

$k = (2\pi f \sqrt{m})^2$

$k = 4\pi^2 f^2 m$

$k = (4) (\pi)^2 (1) (35*10^{-3})$

$k = 1.38N/m$

Therefore the spring constant is 1.38N/m

7 0

3 years ago

What does the atomic number of an element represent?

Zinaida [17]

The number of protons in an atom is represented by the atomic number.

hope this helps

7 0

3 years ago

Which formula describes acceleration?<br><br> m/s^2<br><br><br> m/s<br><br><br> s/m<br><br><br> m2

Aleks04 [339]

Answer:

m/s^2

Explanation:

Force = mass × acceleration

kgm/s^2 = kg × acceleration

where acceleration = Force ÷ mass

= kg m/s^2 ÷ kg

:Acceleration = m/s^2

3 0

3 years ago

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = $\frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm$

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

3 0

3 years ago