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4 years ago

What is sedimentary rock?

2 answers:

3 0

<span>Sedimentary rocks are often deposited in layers, and frequently contain fossils

hope this helps best of luck :)
</span>

9966 [12]4 years ago

3 0

Hey!

Here's a simple definition for Sedimentary Rock:

A Sedimentary Rock is a rock formed of mechanical, chemical, or organic sediment.

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8. Una fuerza de 100 N actúa sobre un cuerpo que se desplaza a lo largo de

vitfil [10]

Answer:

50m

Explanation:

8 0

4 years ago

Electrons are continuously being knocked out of air molecules by cosmic ray particles from space. Once the electrons are free, t

nata0808 [166]

(a) $1.25\cdot 10^{-14} J$

The change in potential energy of the electron is given by:

$\Delta U=q E d$

where

$q=1.6\cdot 10^{-19}C$ is the magnitude of the electron's charge

$E=150 N/C$ is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

$\Delta U=(1.6\cdot 10^{-19}C)(150 N/C)(520 m)=1.25\cdot 10^{-14} J$

(b) 78 kV

The potential difference the electron has moved through is given by

$\Delta V=Ed$

where

$E=150 N/C$ is the magnitude of the electric field

d = 520 m is the distance through which the electron has moved

Substituting into the equation, we find

$\Delta V=Ed=(150 N/C)(520 m)=78,000 V=78 kV$

4 0

3 years ago

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

4 0

3 years ago

A 40.0 kg box is placed on a 2.00 m tall shelf. If the box falls off the shelf, what is its kinetic energy as it strikes the gro

RUDIKE [14]

M = 40 Kg , g=9.8 m/s² , h = 2 m

PE = m g h

PE = (40) (9.8) (2)

PE = 784 J

KE = PE

½m v² = m g h

½ v² = g h

½ v² = (9.8) (2)

½ v² = 19.6

v² = 19.6×2

v² = 39.2

V = √39.2

V = 6.26 m/s

KE = ½mv²

KE = ½(40) (6.26)²

KE =783.8 J

6 0

3 years ago