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3 years ago

1. A body whose mass is 2 kg and has a volume of 500cm just floats

Physics

1 answer:

GarryVolchara [31]3 years ago

8 0

Answer:

Density equals mass over volume so 2 divided 500

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Which is NOT a property of gold? Select one: a. rare b. malleable c. tarnishes d. corrosion resistant

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If I'm correct, gold does not tarnish so C.

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4 years ago

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In an experiment, the explanation of the expected outcome, based on research, is the _______________.

vladimir1956 [14]

C. Hypothesis
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3 years ago

The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward

ch4aika [34]

The object's <u>initial velocity</u> is equal to $-10.59\frac{m}{s}$

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

$y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}$

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

So, isolating the initial velocity, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}$

Finally, substituting and calculating, we have:

$v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}$

Hence, we have that the <u>initial velocity</u> of the object is $-10.59\frac{m}{s}$

Have a nice day!

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4 years ago

If an object is dropped from a tall building and hits the ground 3.0 s. later, what is the magnitude of the object's displacemen

marin [14]

Find the velocity of the object after one second.

v = vo + at

v = (0 m/s) + (9.8 m/s^2)(1 s)

v = 9.8 m/s

Now, using that, you can find the displacement in that one second between 1 and 2.

d = vot + (1/2)at^2

d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2

d = 14.7 m

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3 years ago

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A piano string having a mass per unit length equal to 5.20 10-3 kg/m is under a tension of 1 450 N. Find the speed with which a

DedPeter [7]

Answer:

The wave in the string travels with a speed of 528.1 m/s

Explanation:

Wave speed of sound waves in a string, v, is related to the Tension in the string, T, and the mass per unit length, μ, by the relation,

v = √(T/μ)

μ = 5.20 × 10⁻³ kg/m

T = 1450N

v = √(1450/0.0052) = 528.1 m/s

Hope this Helps!!!

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3 years ago