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3 years ago

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A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break

s as the elevator accelerates. What was the magnitude of the elevator's minimum acceleration? and What was the direction of the elevator's minimum acceleration?

1 answer:

Varvara68 [4.7K]3 years ago

8 0

Answer:

$2.4 m/s^2$

Downward

Explanation:

We are given that

Mass of monkey=12.8 kg

Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

$T=m(g+a)$

Where $g=9.8 m/s^2$

Substitute the values

$156=12.8(a+9.8)$

$9.8+a=\frac{156}{12.8}=12.2$

$a=12.2-9.8=2.4 m/s^2$

Hence, the acceleration = $a=2.4 m/s^2$

Direction of the elevator's minimum acceleration is downward because the elevator moves downwards.

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How energy is converted from one form to other

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Explanation:

Energy can be converted from one form to another. Examples: Gasoline (chemical) is put into our cars, and with the help of electrical energy from a battery, provides mechanical (kinetic) energy. ... Similarly, purchased electricity goes into an electric bulb and is converted to visible light and heat energy.

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I need answers and solvings to these questions

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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

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3 years ago

A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?

posledela

Answer:

$118\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of $v= 295\; \rm m\cdot s^{-1}$ . In other words, the wave would have traveled $295\; \rm m$ in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that $295\; \rm m$ ? The wavelength of this wave $\lambda = 2.50\; \rm m$ gives the length of one wave cycle. Therefore:

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That is: there are $118$ wave cycles in $295\; \rm m$ of this wave.

On the other hand, Because that $295\; \rm m$ of this wave goes through that point in each second, that $118$ wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

$f = 118\; \rm s^{-1} = 118\; \rm Hz$ .

The calculations above can be expressed with the formula:

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

where

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Answer:

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Explanation:

It is given that,

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