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Flura [38]
3 years ago
14

Look at the image of this rock formation near the coast of Palau, Italy. Which factors could contribute to a collapse of the led

ge? Select all that apply.
In this photograph, a weathered boulder sits on top of a cliff. The rocks looks like it is covered in craters and large pieces are missing.

a. wind
b. water
c. ice
d. gravity
Physics
2 answers:
Katena32 [7]3 years ago
7 0

Answer: The answers are A. Wind, B. Water, and D. Gravity.

Explanation: All four factors contribute to weathering and erosion; however, ice is not shown in the image, so it likely did not play a role in the collapse of the cliff.

(Next time please provide an image so it would be easier!)

enyata [817]3 years ago
4 0

The answer is wind and gravity only. Thank me later.

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<span>When two point charges are a distance d apart, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed to When two point charges are a distance d apart, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed to d/âš2</span>
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3 years ago
Ryan swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill on him:
nexus9112 [7]

Part 1

If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position

F_g = ma

mg = m\frac{v^2}{R}

g = \frac{v^2}{R}

v = \sqrt{Rg}

given that

R = 1 m

g = 9.8 m/s^2

now from above equation we have

v = \sqrt{1(9.8)} = 3.13 m/s

Part b)

for minimum value of angular speed we will have

\omega = \frac{v}{R}

\omega = \frac{3.13}{1}

\omega = 3.13 rad/s

3 0
3 years ago
Helppp<br> In Static Electricity, The charges do not ____
Anna11 [10]

Answer:

basically I will tell you the definition

Explanation:

so when charges are unbalanced statistic energy is formed positive attract negative and negative attracts positive like repell while unlike attract .

5 0
3 years ago
What is the frequency of a wave that has a period of vibration of 2 seconds?
34kurt

Answer:

The answer is 0.5 Hz

Explanation:

Its pretty easy to get the answer. One hertz (Hz) is equal to one cycle or period per second. So, just divide the period by the number of seconds.

1 period/2 secs = 1/2 Hz or 0.5 Hz

7 0
3 years ago
What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

5 0
3 years ago
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