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zzz [600]
3 years ago
5

1 m is equivalent to how many centimeters in the metric system

Physics
1 answer:
gladu [14]3 years ago
5 0

100. A centimeter is a 100th of a meter.

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Which tools would be necessary to determine whether or not a large block will float, without using water?
iren2701 [21]

Answer:

The answer is A Ruler and Balance

Explanation:

6 0
3 years ago
I don’t understand I need help asap
hammer [34]
For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s
3 0
3 years ago
The average depth of Indian Ocean is about 3000m.Calculate the fractional compression. ▲v/v, of water at the bottom of the ocean
Klio2033 [76]

Answer:

∴ fractional compression = 1.34 × 10⁻²

Explanation:

given,

depth of Indian ocean = 3000 m

Bulk modulus of the water = 2.2 x 10⁹ N/m²

We know,

P = P₀ + ρgh

P₀ is the atmospheric pressure

P₀ = 10⁵ N/m²

ρ is the density of the water, 1000 Kg/m³

P = 10⁵ + 1000 × 9.8 × 3000 = 2.94 × 10⁷ N/m²

using formula,

B = P/{-∆V/V}

B is bulk modulus and { -∆V/V} is the fractional compression

\dfrac{-\Delta V}{V} = \dfrac{2.94 \times 10^7}{2.2 \times 10^9}

\dfrac{-\Delta V}{V} =1.34 \times 10^{-2}

∴ fractional compression = 1.34 × 10⁻²

7 0
3 years ago
What is the acceleration of the train
Zepler [3.9K]

Answer:

a=1.94 ms−2. is acceleration of the train

5 0
2 years ago
You are hiking along a trail in a wide, dry canyon where the outdoor temperature is T = 29.5° C. To determine how far you are aw
STALIN [3.7K]

Answer:

<em>The canyon is approximately 314 meters away</em>

Explanation:

<u>Speed of Sound</u>

If we emit sounds in an open space where a large obstacle (like a mountain) is expected to return the sounds, then it will travel forth and back at a given speed for a certain time. We can assume the speed of sound is constant, so we could know the approximate distance of the mountain (or canyon in our case) by the known formula.

x=v_st

Where v_s is the speed of sound and t is half the time we hear our echo.

The speed of sound in m/s can be calculated from the approximate formula in terms of the temperature T in degrees Celsius

v_s=331.3+0.606T

We have T=29.5^oC, so

v_s=331.3+0.606(29.5)

v_s=349.177\ m/s

Let's compute x, for t=1.8/2=0.9 seconds

x=v_st=(349.177)(0.9)

x=314.26\ m

The canyon is approximately 314 meters away

7 0
3 years ago
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