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JulijaS [17]
3 years ago
13

A machine lifts a 1000 N object 40 meters in 15 seconds. How much work is done by the machine

Physics
1 answer:
patriot [66]3 years ago
6 0

Answer:

40000 J or 40 kJ

Explanation:

Work is the net force times the distance.

W = Fd

W = (1000 N) (40 m)

W = 40000 J

Notice that the time doesn't affect the amount of work done.

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A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
DENIUS [597]

Answer:

   C = 4,174 10³ V / m^{3/4} ,  E = 7.19 10² / ∛x,    E = 1.5  10³ N/C

Explanation:

For this exercise we can calculate the value of the constant and the electric field produced,

Let's start by calculating the value of the constant C

           V = C x^{4/3}

           C = V / x^{4/3}

            C = 220 / (11 10⁻²)^{4/3}

            C = 4,174 10³ V / m^{3/4}

To calculate the electric field we use the expression

            V = E dx

             E = dx / V

             E = ∫ dx / C x^{4/3}

            E = 1 / C  x^{-1/3} / (- 1/3)

            E = 1 / C (-3 / x^{1/3})

We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E

            E = 3 / C     (0- (-1 / x^{1/3}))

            E = 3 / 4,174 10³   (1 / x^{1/3})

           E = 7.19 10² / ∛x

for x = 0.110 cm

          E = 7.19 10² /∛0.11

          E = 1.5  10³ N/C

6 0
4 years ago
Physics. Need help. Brainlieast answer for most/ all of the answers answered
Mumz [18]

<u>ALL of the following work assumes NO AIR RESISTANCE:</u>

1). an object moving under the influence of only gravity, and not in orbit;  its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both  hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds;   b). 13.1 meters;   c). 17.6 m/s down;   d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s;  b). 29.7 m/s

10). 7.24 meters

11).  700 meters

12).  A). 103.7 meters ( ! she's in big trouble ! );     B).  17.5 meters

3 0
3 years ago
L'aigua es una mescla de d'oxigen i hidrogen
pshichka [43]

Answer:

Una Mezcla Homogénea es aquella mezcla en la que las sustancias que la forman poseen una combinación uniforme.Son ejemplos de Mezclas Homogéneas: Compuesta

Explanation:

Aire (es una mezcla de gases homogénea formada principalmente por de nitrógeno, oxígeno, vapor de agua, dióxido de carbono...)

Leche (mezcla de agua, carbohidratos, proteínas...)

Bebida alcohólica (mezcla de agua y alcohol etílico)

Acero (mezcla de elementos aleados como el hierro, el carbono y otras sustancias)

Petróleo (mezcla de hidrocarburos)

Agua de mar (mezcla de agua, cloruro sódico y otras sustancias)

 

Mezcla de agua y sal disuelta

Agua azucarada (mezcla de agua y azúcar)

Aleación metálica (las aleaciones metálicas son mezclas en las que se combinan diferentes metales de una manera homogénea y definida)

Perfume (mezcla de agua y otras sustancias olorosas cuya composición es uniforme)

3 0
3 years ago
Can anyone help me with this??? Anyone familiar with this sorta project??
SIZIF [17.4K]

yah set up an experiment do u have the rocks with u?


7 0
3 years ago
Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

8 0
3 years ago
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