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3 years ago

Which phrase best descirbes how scientists use the data they collect

Physics

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

Please give the context of questions

Explanation:

It really helps people answer your questions. thanks

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

melisa1 [442]

Answer:

$H_2 = 91.55 km$

Explanation:

Gravity on the surface of planet is given as

$g = \frac{GM}{R^2}$

as we know that

$M = 8.93 \times 10^{22} kg$

$R = 1821 km$

now gravity on the planet is

$g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}$

so we have

$g = 1.8 m/s^2$

now we know that

$H_{max} = \frac{v^2}{2g}$

so we will say

$\frac{H_1}{H_2} = \frac{g_2}{g_1}$

$\frac{500}{H_2} = \frac{9.81}{1.8}$

$H_2 = 91.55 km$

5 0

3 years ago

A stone that starts at rest free falls for 7.0 s. How far does the stone fall in this time?

vredina [299]

Gravitational acceleration is approx 9.8 m/s
Time is 7s

a=9.8 m/s
t=7s

a = d/t^2

therefore:

d = a * t^2

d = 9.8 * 7^2

d = 9.8 * 49

d = 480.2 [m]

7 0

3 years ago

A ball is dropped from a height of 23.5 meters. Assuming no air resistance, how many seconds will it take the ball to hit the

Alex Ar [27]

Answer:

About 2.19 seconds

Explanation:

$d=v_ot+\dfrac{1}{2}at^2$

Since the ball is dropped from rest, there is no initial velocity, and you can write the following equation:

$23.5=\dfrac{1}{2}(9.8)t^2 \\\\t^2\approx 4.79 \\\\t\approx 2.19s$

Hope this helps!

8 0

3 years ago

Read 2 more answers

A block weighing 3.7 kg is suspended from the ceiling of a truck trailer by a hanging bungee cord. The cord has a cross-sectiona

sweet [91]

Answer:

$Y = 2.27 \times 10^{10} N/m^2$

Explanation:

Natural length of the string is given as

$L_o = 43 cm$

length of the string while block is hanging on it

$L = 53 cm$

extension in length is given as

$\Delta L = 10 cm$

now we have strain in the string is given as

$strain = \frac{\Delta L}{L}$

$strain = \frac{{10 cm}{43 cm}$

$strain = 0.23$

similarly we will have cross-sectional area of the string is given as

$A = 40 \times 10^{-6} m^2$

now the stress in the string is given as

$Stress = \frac{T}{A}$

$Stress = \frac{mg}{A}$

$Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}$

$stress = 9.07 \times 10^5 N/m^2$

Now Young's Modulus is given as

$Y = \frac{stress}{strain}$

$Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}$

$Y = 2.27 \times 10^{10} N/m^2$

5 0

3 years ago

3) Using the rating of perceived exertion scale that we

Schach [20]

Answer:

What do you mean

Explanation:

7 0

3 years ago