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3 years ago

Two boxes are connected by a string. The first box has a mass of 4.2 kg and the

Physics

1 answer:

denis23 [38]3 years ago

8 0

Answer:

........

Explanation:

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A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The

Ber [7]

Answer:

1. $v=14.2259\ m.s^{-1}$

2. $F_T=25.8924\ N$

3. $\lambda=29.6373\ m$

Explanation:

Given:

mass of slinky, $m=0.87\ kg$

length of slinky, $L=6.8\ m$

amplitude of wave pulse, $A=0.23\ m$

time taken by the wave pulse to travel down the length, $t=0.478\ s$

frequency of wave pulse, $f=0.48\ Hz=0.48\ s^{-1}$

$\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel$

$v=\frac{6.8}{0.478}$

$v=14.2259\ m.s^{-1}$

<em>Now, we find the linear mass density of the slinky.</em>

$\mu=\frac{m}{L}$

$\mu=\frac{0.87}{6.8}\ kg.m^{-1}$

We have the relation involving the tension force as:

$v=\sqrt{\frac{F_T}{\mu} }$

$14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }$

$202.3774=F_T\times \frac{6.8}{0.87}$

$F_T=25.8924\ N$

We have the relation for wavelength as:

$\lambda=\frac{v}{f}$

$\lambda=\frac{14.2259}{0.48}$

$\lambda=29.6373\ m$

8 0

3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago

When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo

Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And $\theta$ be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim . Both of their motions are subjected to parallel component of the gravitational acceleration $gsin\theta$

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is $s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8$

As both of them cover the same distance

$45000a = 3at_J^2/8$

$t_J^2 = 45000*8/3 = 120000$

$t_J = \sqrt{120000} = 346.4 s$

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0

4 years ago

The atomic mass of an atom do not contain electrons because

Dimas [21]

Answer:

Electrons are so small that it does not affect the mass of atom .

Explanation:

Electrons are much smaller in mass than protons, weighing only 9.11 × 10^-28 grams, or about 1/1800 of an atomic mass unit. Therefore, they do not contribute much to an element's overall atomic mass.

7 0

3 years ago

Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u

AnnZ [28]

Answer:

#See solution for details.

Explanation:

Tools: $stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring \ device$ .

$Experiment \ 1$ :Calculate the speed of the wave using the time, $t$ it takes to travel along the rope. Rope's length, $L$ is measured using the meter stick.

-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

-Speed of the pulse can then be obtained as:

$v=\frac{L}{t}$

$Experiment \ 2$ : Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

$v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}$

-Use the measuring stick and measuring scale to determine $L,m$ values of the rope then obtain $\mu$ .

-Use the force measuring constant to determine $F$ . These values can the be substituted in $Experiment \ 1$ to obtain $v.$

4 0

3 years ago