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3 years ago

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A 2-m3 rigid tank initially contains air whose density is 1.18 kg/m3 . The tank is connected to a high-pressure supply line thro

ugh a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 5.30 kg/m3 . Determine the mass of air that has entered the tank?

1 answer:

sdas [7]3 years ago

5 0

Explanation:

It is known that density is the mass present in per unit volume.

Mathematically, Density = $\frac{mass}{volume}$

Since, it is given that $d_{1}$ is 1.18 $kg/m^{3}$ , $d_{2}$ is 5.30 $kg/m^{3}$ , and volume is 2 $m^{3}$ .

Therefore, mass of air that has entered will be $m_{2}$ - $m_{1}$ and it will be calculated as follows.

$d_{2} - d_{1}$ = $\frac{m_{2} - m_{1}}{Volume}$

$m_{2} - m_{1}$ = $(5.30 kg/m^{3} - 1.18 kg/m^{3}) \times 2 m^{3}$

= 8.24 kg

Thus, we can conclude that mass of air that has entered the tank is 8.24 kg.

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The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

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Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It is neccesary calculate the force:

$F=(m+m)*g*sin\theta$

Here, g = gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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3 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

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3 years ago

A motorbike is travelling with a velocity of 3m/s. It accelerates at a rate of 9.3m/s for 1.8s. Calculate the distance it travel

Misha Larkins [42]

Explanation:

Consider the kinematic equation,

$x = vt + \frac{1}{2} a {t}^{2}$

where x is the distance traveled, v is the initial velocity, a is the acceleration and t is time. By plugging in known values and solving for x,

$x = 3(1.8) + \frac{1}{2} (9.3) {1.8}^{2}$

through simple algebra we get

$x = 20.466$

where this is the distance traveled in meters.

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what does the term equilibrium refer to? a. the resting position of the wave b. the highest point in the wave c. the lowest poin

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A. the resting position of a wave

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A hammer taps on the end of a 3.4-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pu

OLEGan [10]

Answer:

S = 2266.67 m/s

Explanation:

Given,

length of the metal = 3.4 m

pulses are separated in time = 8.4 ms

speed of sound in air= 343 m/s

speed of sound in this metal = ?

time taken

$t = \dfrac{distance}{speed}$

$t = \dfrac{3.4}{343}$

t = 9.9 ms

speed of sound in the metal is fast

t = 9.9 - 8.4 = 1.5 ms

time for which sound is in metal is equal to 1.5 ms

speed of sound in metal

$speed= \dfrac{distance}{time}$

$S = \dfrac{3.4}{1.5 \times 10^{-3}}$

S = 2266.67 m/s

Speed of sound in metal is equal to S = 2266.67 m/s

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3 years ago