Question 7 of 20

Una roca sobre una superficie de hielo está siendo tirada horizontalmente por tres cuerdas como se ve en la figura, donde F=95N,

KatRina [158]

Hablas español??? or portuguese

6 0

3 years ago

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

3 years ago

How do prokaryotic cells replicate?

Alika [10]

Answer:

The usual method of prokaryote cell division is termed binary fission. The prokaryotic chromosome is a single DNA molecule that first replicates, then attaches each copy to a different part of the cell membrane. When the cell begins to pull apart, the replicate and original chromosomes are separated.

6 0

3 years ago

An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600

dmitriy555 [2]

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

4 0

4 years ago

A sound from a source has an intensity of 270 dB when it is 1 m from the source.

Rufina [12.5K]

Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{2}$ is the intensity at distance 2
$I_{1}$ is the intensity at distance 1
$d_{2}$ is distance 2
$d_{1}$ is distance 1

We can infer for our problem that $I_{1}=270$ , $d_{1}=1$ , and $d_{2}=3$ . Lets replace those values in our formula to find $I_{2}$ :
$\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
$\frac{I_{2} }{270} =( \frac{1}{3} )^2$
$\frac{I_{2} }{270} = \frac{1}{9}$
$I_{2}= \frac{270}{9}$
$I_{2}=30$ dB

We can conclude that the intensity of the sound when is <span>3 m from the source is 30 dB.</span>

8 0

3 years ago