Ask question

Ask question

All categories

vladimir1956 [14]

2 years ago

6

For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and sto

pped in a 1.2-ms-long encounter with a hard floor?

1 answer:

arsen [322]2 years ago

6 0

Answer:

Acceleration will be equal to $3689.16m/sec^2$

Explanation:

We have given initial height h = 1 m

Time instant t = 1.2 m-sec = 0.0012 sec

Initial velocity u = 0 m/sec

From third equation of motion we know that $v^2=u^2+2gh$

So $v^2=0^2+2\times 9.8\times 1$

v = 4.427 m/sec

We have to find the acceleration

Acceleration is time rate of change of velocity

So acceleration $a=\frac{v}{\Delta t}=\frac{4.427}{0.0012}=3689.16m/sec^2$

So acceleration will be equal to $3689.16m/sec^2$

You might be interested in

Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.40 kg and is initially moving to t

Dmitrij [34]

Answer:

The velocity of the first block is 1.15m/s while of the second block 2.56m/s.

Explanation:

Momentum is only conserved in an isolated system, and because this problem requires us to find the value of the two variables, we need two equations; therefore, to conserved momentum the energy must be released in to the system only after the collision has occurred.

Therefore, from conservation of momentum

$m_1v_1= m_1v_{1f}+m_2v_{2f}$

$0.4(2.4)= 0.4v_{1f}+0.5v_{2f}$

$\boxed{\:0.96= 0.4v_{1f}+0.5v_{2f}}$

and from conservation of energy

$\frac{1}{2}m_1v_1^2+1.2 = \frac{1}{2} m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2$

$m_1v_1^2+2.4 = m_1v_{1f}^2+m_2v_{2f}^2}$

$\boxed{\:3.804= 0.4v_{1f}^2+0.5v_{2f}^2.}$

Thus, we have two equations and two unknowns

$(1). \: 0.4v_{1f}+0.5v_{2f}=0.96$

${(2). \: 0.4v_{1f}^2+0.5v_{2f}^2 =3.804$

which has solutions

$(v_{1f}, v_{2f}) = (1.15,2.56)$

and

$(v_{1f}, v_{2f}) = (1.15,-2.56)$

Since the blocks cannot pass through each other, the 0.5kg block cannot have $v_{2f} = -2.56m/s$ (moves to the left) while the 0.4 kg block has $v_{1f} = 1.15m/s$ (moves to the right); therefore, we take the first solution for the velocities:

$(v_{1f}, v_{2f}) = (1.15,2.56)$ .

Thus , the velocity of the first block is 1.15m/s while of the second block 2.56m/s.

5 0

2 years ago

Read 2 more answers

What are the dark areas on the surface of the Sun?

Nitella [24]

Sunspots,despite being called sunspots they are still thousands of degrees hot with an average cycle of 11years per minimum and the climax of these phenomenon s

4 0

3 years ago

Read 2 more answers

Why do you think festival dances should be introduced to the younger generations of today

Nimfa-mama [501]

Answer:

Festival dances should be introduced to the younger generations of today because they teach them about different cultures. Festival dances are also a way to be active and have a fun time.

Explanation:

4 0

2 years ago

Read 2 more answers

Two tectonic plates moving toward one another are at a

nordsb [41]

Answer:

A. cause i just took the test

Explanation:

6 0

2 years ago

Read 2 more answers

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick

rewona [7]

Answer:

The new time period is $T_2 = 3.8 \ s$

Explanation:

From the question we are told that

The period of oscillation is $T = 5 \ s$

The new length is $l_2 = 0.76 \ m$

Let assume the original length was $l_1 = 1 m$

Generally the time period is mathematically represented as

$T = 2 \pi \sqrt{ \frac{ I }{ mgh } }$

Now I is the moment of inertia of the stick which is mathematically represented as

$I = \frac{m * l^2 }{12 }$

So

$T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }$

Looking at the above equation we see that

$T \ \ \ \alpha \ \ \ l$

=> $\frac{ T_2 }{T_1} = \frac{l_2}{l_1}$

=> $\frac{ T_2}{5} = \frac{0.76}{1}$

=> $T_2 = 3.8 \ s$

3 0

3 years ago