Ask question

Ask question

All categories

julia-pushkina [17]

2 years ago

14

the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

le shown above is most nearly

1 answer:

JulsSmile [24]2 years ago

5 0

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

<em>Mass of the first block, m₁ = 4.0 kg</em>
<em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

<em>T </em><em>is the tension in the connecting string due weight of the first block</em>

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

You might be interested in

A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina

elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car $a=1.6m/sec^2$ in negative direction so acceleration will be $a=-1.6m/sec^2$

From first equation of motion we know that

v = u+at

So $v=0+(-1.6)\times 5.8=-9.28m/sec$

So final velocity will be -9.28 m/sec

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:

<h2>In the x-axis: </h2>

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

<h2>In the y-axis: </h2>

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (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 Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will _____

Bingel [31]

When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will <u>"pull back".</u>

Newton's third law of motion expresses that, at whatever point a first question applies a power on a second object, the first object encounters a power meet in extent however inverse in heading to the power that it applies.

Newton's third law of movement reveals to us that powers dependably happen in sets, and one question can't apply a power on another without encountering a similar quality power consequently. We once in a while allude to these power matches as "action-reaction" sets, where the power applied is the activity, and the power experienced in kind is the response (despite the fact that which will be which relies upon your perspective).

6 0

3 years ago

Read 2 more answers

Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 4.0 mm

loris [4]

Answer:

4.4721m

Explanation:

#Use Pythagorean theorem to find the distance between the two speakers:

$a^+b^2=c^2\\4^2+2^2=d^2\\\sqrt{20}\ mm=d$

There are antinodes 1/4,1/2 and 3/4 of the distance between speakers.

The greatest antinode is 3/4-1/4=1/2

#Distance between consecutive antinodes is:

$0.5\times \sqrt{20}=\lambda/2\\\\\lambda=4.4721m$

Hence, the maximum possible wavelength of the sound waves is 4.4721m

6 0

3 years ago

when 10 similar coins are dropped into a graduated cylinder from 75 ml to 100ml. what is average volume of each coin?

Alex787 [66]

Explanation:

Volume of 10 coins = 100ml - 75ml = 25ml

Volume of 1 coin = 25ml / 10 = 2.5ml

The average volume of each coin is 2.5ml.

8 0

2 years ago

Read 2 more answers