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melamori03 [73]

2 years ago

5

Which is not true of friction? A. Causes wear and tear of the surfaces B. Helps us to fall easily on roads C. Produces heat D. P

roduces light E. Stops bodies when they move over each other.

2 answers:

Ganezh [65]2 years ago

5 0

Answer:

B.

Hope this helps

Aleks04 [339]2 years ago

3 0

The answer is B

i need 20 works

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Describa con sus palabras que fue lo que descubrio galileo en su legendario experimento en la torre inclinada de pisa

german

Answer:

En 1589 Galileo realizó un experimento lanzando dos bolas de diferentes masas desde la famosa Torre Inclinada de Pisa para demostrar que el tiempo de caída es independiente de la masa de la bola. A través de este experimento, Galileo descubrió que los cuerpos caían casi simultáneamente, refutando la teoría de Aristóteles de que la tasa de caída era proporcional a la masa del cuerpo.

Debido a la imperfección de los equipo de medición de esa época, la caída libre de los cuerpos era casi imposible de estudiar. En busca de una forma de reducir la velocidad de movimiento, Galileo reemplazó la caída libre por rodar sobre una superficie inclinada, donde había velocidades y resistencia del aire significativamente más bajas. Se notó que con el tiempo, la velocidad del movimiento aumenta: los cuerpos se mueven con aceleración. Se concluyó que la velocidad y la aceleración no dependen ni de la masa ni del material de la pelota.

5 0

3 years ago

RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s

Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, $R=50\ K\Omega=5\times 10^4\ \Omega$

Capacitance, $C=2.1\ \mu F=2.1\times 10^{-6}\ F$

We need to find the expected time constant for this RC circuit. It can be calculated as :

$\tau=R\times C$

$\tau=5\times 10^4\times 2.1\times 10^{-6}$

$\tau=0.105\ s$

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0

2 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

20. Using the picture, how many neutrons does lithium have?

Igoryamba

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

<u>No. of Neutrons = 3 </u>

8 0

3 years ago

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

4 0

2 years ago