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Vlad [161]
3 years ago
5

In the bottom of death valley, you will find layers of gravel deposited by rivers. based on materials presented in class, what i

s a likely explanation for this occurrence of river gravels in the valley bottom?
Physics
1 answer:
nikklg [1K]3 years ago
5 0
<span>In the bottom of death valley, you will find layers of gravel deposited by rivers. based on materials presented in class, the explanation for this occurrence of river gravels in the valley bottom is that t</span>he valley was dropped relative to the mountains by faulting, and rivers now are carrying gravels down from the mountains into the valley and depositing the gravels at the valley bottom.
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A car is moving down the street at 53km/h. A child suddenly runs into the street. If it takes the driver 0.75 sec to react and a
BARSIC [14]

Answer:

11.04m

Explanation:

Convert kilometers to meters: 53km/h x 1000 = 53,000m/h

Convert hours to seconds: 53m/h / 3600 =14.72m/s

14.72 (.75) = 11.04

8 0
3 years ago
Microwave ovens emit microwave energy with a wavelength of 13.0 cm. what is the energy of exactly one photon of this microwave r
zysi [14]
First, let's calculate the frequency of this radiation, which is given by:
f= \frac{c}{\lambda}
where c is the speed of light and \lambda is the photon wavelength. For this radiation, photons have wavelength of
\lambda=13.0 cm=0.13 m
Therefore their frequency is
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{0.13 m}=2.3\cdot 10^9 Hz

The energy of a photon with frequency f is given by
E=hf
where h is the Planck constant. By using the frequency we found before, we find the energy of a single photon of this radiation:
E=hf=(6.6 \cdot 10^{-34} Js)(2.3 \cdot 10^9 Hz)=1.52 \cdot 10^{-24} J
4 0
4 years ago
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The allele for a hybrid tall pea plant is represented as
koban [17]
Hey there! 

The allele for a hybrid tall pea plant is represented as Tt

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4 years ago
A meter stick swinging from one end oscillates with a frequency f0. What would be the frequency, in terms of f0 , if the bottom
snow_lady [41]

To solve this problem we will proceed to define the Period of a stick, then we will define the frequency, which is the inverse of the period. We will compare the change suffered by the new length and replace that value. The Time period of meter stick is

T = 2\pi\sqrt{ \frac{L}{g}}

Here,

L = Length

g = Gravity

At the same time the frequency is

f = \frac{1}{T}

Therefore the frequency in Terms of the Period is

f_0 = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

If bottom third were cut off then the new length is

L' = \frac{2}{3} L

Replacing this value at the new frequency we have that,

f ' = \frac{1}{2\pi} \sqrt{\frac{3g}{2L}}

f' = \sqrt{\frac{3}{2}}(\frac{1}{2\pi}\sqrt{\frac{g}{L}})

Finally,

\therefore f' = \sqrt{\frac{3}{2}}f_0

3 0
3 years ago
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Partial products for 56* 31
stiks02 [169]
56 * 31=1,736. Hope this answers your question
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