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Zolol [24]
4 years ago
13

What is sedimentary rock?

Physics
2 answers:
Margaret [11]4 years ago
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. 
You might be interested in
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+1540=t^{2} -28.75t+154

To solve this equation we use the following formulas

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

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)}=21.63st=\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
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