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3 years ago

Can someone help me this question

Physics

1 answer:

vladimir2022 [97]3 years ago

4 0

Answer:

Volume of wood is 68-50 = 18 cm³

Explanation:

<h2>Hope it helps you</h2>

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Which method of popcorn popping transfers heat into the kernels without any direct

Ivanshal [37]

Answer:

you plug in and turn on the popper, and then the hot air begins to rise in the popper while cooler air falls. As hot air circulates past the popcorn kernels and so the kernels absorb the hear

8 0

2 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0

3 years ago

at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g

Readme [11.4K]

The mass of water that must be raised is $5.25\cdot 10^7 kg$

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

$P_{out} = 0.70 P_{in}$

where $P_{in}$ is the power in input.

The power in input can be written as

$P_{in} = \frac{W}{t}$

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

$W=mgh$

where

m is the mass of water

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

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

$P_{out} = 0.70 \frac{mgh}{t}$

Where

$P_{out} = 150 MW = 150\cdot 10^6 W$

And solving for m, we find:

$m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg$

Learn more about power:

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3 0

3 years ago

If the height of the ramp was 1.2 m above the floor, how long would it take for the marble to hit the ground after it left the r

Evgesh-ka [11]

It would take at less 10 minte i guess this the right awnser

8 0

3 years ago

When measuring an unknown voltage with an analog VOM, you should first

Soloha48 [4]

If you have no idea what the voltage is that you're about to measure,
then you should set the meter to the highest range before you connect
it to the two points in the circuit.

Analog meters indicate the measurement by moving a physical needle
across a physical card with physical numbers printed on it. If the unknown
voltage happens to be 100 times the full range to which the meter is set,
then the needle may find itself trying to move to a position that's 100 times
past the highest number on the meter's face. You'll hear a soft 'twang',
followed by a louder 'CLICK'. Then you'll wonder why the meter has no
needle on it, and then you'll walk over to the other side of the room and
pick up the needle off the floor, and then you'll probably put the needle
in your pocket. That will end your voltage measurements for that day,
and certainly for that meter.

Been there.
Done that.

8 0

3 years ago

Read 2 more answers

Can someone help me this question ​

Can someone help me this question