Se deja caer una caja de madera de 4.5kg de masa desde una altura de 2.25metros .

The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one

ozzi

Answer:

e. 400 Hz

Explanation:

In closed organ pipe, only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz , 1800 Hz .

Frequency not possible is 400 Hz .

7 0

3 years ago

An experiment is performed on an unknown material and produces the given heat curve. The temperature of the material is shown as

BlackZzzverrR [31]

Answer:

In the table, 1=46.7 °C, 1=165 J, 2=819 J, 3=1510 J, and 4=2830 J.

Other experiments determine that the material has a temperature of fusion of

fusion =235 °C and a temperature of vaporization of vapor=481 °C.

If the sample of material has a mass of =8.60 g, calculate the specific heat when this material is a solid, and when it is liquid, l

4 0

3 years ago

How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half

beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

$V=RI$

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

$I=\frac{V}{R}$

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: $V'=\frac{V}{2}$

So, the new current will be

$I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}$

so, the current will drop to half of its original value.

4 0

3 years ago

Một oto đang chuyển động thẳng đều với vận tốc 36km/h thì hãm phanh sau 20s thì ô tô dừng lại quãng đường oto đi được là

saw5 [17]

36km/h=10m/s
v=s/t=>s=200m

3 0

3 years ago

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

3 years ago