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IRINA_888 [86]
10 months ago
6

Radio waves travel at a speed of 1.7 times 10^8 m/s through ice. A radio wave pulse sent into the Antarctic ice reflects off the

rock at the bottom and returns to the surface in 32.9 times 10^-6 s. Part A How deep is the ice? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Nostrana [21]10 months ago
4 0

The values with (1/2)(9.8 m/s2) (4 - X)^2 = (1.7 × 10^8 m/s) (X) —> We discover with the calculator X = 4.40 × 10^-7. The depth of the ice is then calculated by multiplying by the speed 74.8 m. Through ice, radio waves move at a speed of 1.7 x 108 m/s.

What is radio waves?

A radio wave pulse that is delivered into the Antarctic ice returns after reflecting off the bottom rock. A radio wave pulse that is transmitted into the Antarctic ice bounces off the subsurface rock and resurfaces. ice in Antarctica Through ice, radio waves move at a speed of 1.7 108 m/s. The rock reflects the radio wave pulse that was transmitted into the Antarctic ice. The amount of time that passed between the signal being sent and being received by the earth station.

To learn more about Antarctic from given link

brainly.com/question/3200132

#SPJ4

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Explain a situation in which you can accelerate even though your speed doesn’t change.
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A few examples:
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4 0
3 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

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(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

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For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

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a is the acceleration

In this problem,

u = 0

t = 10 s

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Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

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x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

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