All categories

3 years ago

A conveyer belt moves a 40 kg at a velocity of 2/ms what is the kinetic energy of the box while it is on the conveyor belt

1 answer:

5 0

KE = (1/2)·(mass)·(speed)²

It doesn't matter whether the object is in a car, on a boat,
falling, on a conveyor belt, or being carried by ants.

KE = (1/2)·(40 kg)·(2 m/s)²

KE = (20 kg)·(4 m²/s²)

KE = 80 kg-m²/s²

KE = 80 Joules

You might be interested in

A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v

IRISSAK [1]

The solution to this task is f00king Dis.

6 0

2 years ago

Twin A makes a one way trip at 0.6c to a star 12 light years away while twin B stays on Earth. Each twin sends the other a signa

hram777 [196]

Answer:

8 signals received by twin A during the trip.

Explanation:

Given that,

Distance = 12 light year

Speed = 0.6 c

Time = 1 year

We need to calculate the time by A

Using formula of time

$T=t\sqrt{\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}}$

Put the value into the formula

$T=1\sqrt{\dfrac{1+0.6}{1-0.6}}$

$T=2\ years$

Similarly,

The expression for distance cover by A

$D=d\sqrt{1-\dfrac{v^2}{c^2}}$

$D=12\sqrt{1-(0.6)^2}$

$D=9.6\ ly$

We need to calculate the time

Using formula of time

$t=\dfrac{D}{v}$

$t=\dfrac{9.6}{0.6}$

$t=16\ years$

We need to calculate the signals received by twin A

Using formula for number of signals

$n=\dfrac{t}{T}$

Put the value into the formula

$n=\dfrac{16}{2}$

$n=8\ signals$

Hence, 8 signals received by twin A during the trip.

7 0

3 years ago

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the

Shalnov [3]

Answer:

x ’= 1,735 m, measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

x_{cm} = 1.2 -1

x_ {cm} = 0.2 m

Σ τ = 0

w₁ 1.2 + mg 0.2 - W₂ x = 0

x = $\frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}$

x = $\frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}$

let's calculate

x = $\frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}$ 2.9 1.2 + 4 0.2 / 8

x = 0.535 m

measured from the pivot point

measured from the far left is

x’= 1,2 + x

x'= 1.2 + 0.535

x ’= 1,735 m

8 0

3 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0

3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago