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3 years ago

14

A velocity vs. time graph starts at 0 and ends at 10 m/s, stretching over a time- span of 15 s with constant acceleration. What

is the object's net displacement?

1 answer:

Alex777 [14]3 years ago

6 0

Answer:

75m

Explanation:

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The unit for energy is the joule (j). If the energy content of an object is 1,000 J, but then later is found to be 900 J, what i

Dennis_Churaev [7]

Answer: the object transferred some of its energy to its surroundings.

Explanation:

5 0

2 years ago

A spring whose spring constant is 270 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

cricket20 [7]

Answer:

0.02585 BTU

Explanation:

Given: Spring constant, k = 270 lbf/in

Initial force, f =100 lbf

Compression, x = 1 in

Work done can be calculated as follows:

$W = \int {(f+kx)} \, dx \\W = fx + \frac{1}{2}kx^2\\W= (100 lbf)(1 in)+ \frac{1}{2}(270 lbf/in)(1 in)^2\\W= 100+135 = 235 lbf in\\W=235 \times 0.00011 BTU = 0.02585 BTU$

6 0

3 years ago

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.

AleksAgata [21]

Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

Explanation:

The frequency of the $n^{th}$ harmonic of a vibrating string of length <em>L, </em>linear density $\mu$ under a tension <em>T</em> is given by the formula:

$f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}$

a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:

$f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz$

b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:

$f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz$

3 0

2 years ago

For every action, there is an equal and opposite reaction.

KiRa [710]

Answer:

This is newton's 3rd law

Explanation:

3 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago