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Lesechka [4]
3 years ago
13

Suppose that you find the volume of all the oceans to be 1.4×109km3 in a reference book. To find the mass, you can use the densi

ty of water, also found in this reference book, but first you must convert the volume to cubic meters. What is this volume in cubic meters?
Physics
2 answers:
Licemer1 [7]3 years ago
8 0

Answer:

Explanation:

Volume of ocean = 1.4×10^9 km3

To m^3,

1.4 × 10^9 km^3 × (1000 m)^3/(1 km)^3

= 1.4 × 10^18 m^3.

Density of water = 1000 kg/m3

Mass = density × volume

= 1000 × 1.4 × 10^18

= 1.4 × 10^21 kg.

PSYCHO15rus [73]3 years ago
4 0

Answer:

V = 1.4\times 10^{18}\,m^{3},m = 1.435\times 10^{21}\,kg

Explanation:

The volume of all oceans is:

V = (1.4\times 10^{9}\,km^{3})\cdot (\frac{1\times 10^{9}\,m^{3}}{1\,km^{3}} )

V = 1.4\times 10^{18}\,m^{3}

Ocean water has a density of 1025\,\frac{kg}{m^{3}}, the mass of all oceans is:

m = (1025\,\frac{kg}{m^{3}} )\cdot (1.4\times 10^{18}\,m^{3})

m = 1.435\times 10^{21}\,kg

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Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

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v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
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The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
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  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
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U= √{2.20×9.8/sin(73)} = 22.5m/s

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Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

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Answer:

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