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RUDIKE [14]
3 years ago
11

Why is the monolith el capitan in yosemite national park more resistant to erosion than other igneous plutons in the park?

Physics
1 answer:
Harman [31]3 years ago
4 0
Because it is intact and unfractured
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What is the wavelength of a sound wave<br> with a speed of 331 m/s and a frequency<br> of 500 Hz?
ludmilkaskok [199]

Answer:

0.777m

Explanation:

The sound wave has a wavelength of 0.773m.

Explanation:

To solve this problem we have to use the wave equation that is given below:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for  

λ :

λ = v f

Let's plug in our given values and see what we get!

λ = 340 m s

440 s − 1

λ = 0.773 m

Hope this helps, Mark as brainliest if u want

4 0
2 years ago
If you average 22.8 meters/min, how many seconds will it take you to travel 467 m?
Agata [3.3K]
Speed = 22.8m/min
= 22.8 / 60
=0.38m/s

time = distance/ speed
= 467/0.38
= 1252.63s
6 0
3 years ago
Read 2 more answers
1) Which is a characteristic of a metal? A) It looks dull. B) It feels brittle. C) It is a good insulator. D) It is a good condu
max2010maxim [7]

D) It is a good conductor of electricity

6 0
3 years ago
Una grúa efectúa un trabajo de 45000 joules en 0.1
Hunter-Best [27]

Answer:

<em>P = 7.5 KW</em>

<em>P = 7500 W</em>

Explanation:

<u>Potencia Mecánica</u>

Es una medida del trabajo realizado por un agente por unidad de tiempo.

Siendo W el trabajo mecánico y t el tiempo necesario para realizarlo, entonces la potencia viene dada por:

\displaystyle P=\frac{W}{t}

La unidad SI de la potencia es el vatio (W). Otra unidad muy común es el kilovatio (KW). 1 KW = 1000 W.

La grúa realiza un trabajo W=45000 Joules en un tiempo de t=0.1 minutos.

Para calcular la potencia en vatios, es necesario llevar el tiempo a segundos:

t = 0.1 min *60 = 6 segundos.

La potencia se calculará ahora:

\displaystyle P=\frac{45000}{6}

P = 7500 W

En kilovatios:

P = 7500 / 1000 = 7.5 KW

P = 7.5 KW

3 0
3 years ago
We have an initially uncharged hollow metallic sphere with radius of 5.0 cm. I place a small object with a charge of +20 µC at t
Advocard [28]

Answer:

d. 1.8E+7 N/C

Explanation:

In order to find the electric field outside the hollow metallic sphere, we can apply Gauss' Law, using a spherical gaussian surface with radius equal to 10 cm from the center of the sphere.

As the electric field must be normal to the surface at any point (no tangential fields can exist in electrostatic conditions) it must be radially pointed. By symmetry, at a same radius, the magnitude of the field must be the same.

As the dA vector is always normal to the surface and aiming outward, the dot product E*dA, can be taken out of the integral, as follows:

E*A = \frac{Qenc}{\epsilon0}

where Qenc, is the total charge enclosed by the gaussian surface. Just due to the conservation of charge, this charge must be equal to +20 μC.

Now, how can this charge be distributed on the outer surface of the shell?

If we apply Gauss´Law to a gaussian surface with a radius just inside the sphere (between the inner and outer surface), we will find that the flux is 0, due to the electric field is 0 inside a conductor.

We could write the same equation as above:

E*A = \frac{Qenc}{\epsilon0} = 0

If the left side of the equation is 0, the right one must be zero too:

⇒ Qenc = 0  ⇒ Qc + Qin = 0 ⇒ Qin = -20 μC.

As the metallic sphere must remain neutral, an equal and opposite charge must build up on the outer surface:

⇒ Qou = +20 μC

The other parameters in the equation are:

r = 0.1 m

ε₀ = 8.85*10⁻¹² C²/N*m²

Replacing by the values, we can solve for E as follows:

E = \frac{1}{4*\pi*\epsilon0} *\frac{Qenc}{r^{2}} = \frac{1}{4*\pi*8.85e-12} *\frac{+20e-6C}{(0.1m)^{2}} = 1.8e7 N/C

⇒ E = 1.8*10⁷ N/C

So, the statement d. is true.

3 0
3 years ago
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