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tensa zangetsu [6.8K]

3 years ago

13

How much potential energy does a 40-kg medicine ball gain when it is lifted 5 m? Use the formula: GPE= mass X gravity (9.8 m/s2)

X Height
Question 16 options:

A 20.4 J

B 100 J

C 1960 J

D 200 J

1 answer:

Ronch [10]3 years ago

4 0

The answer is c: <span>1960 J

</span>Potential Energy :
<span>PE = m x g x h = 40*9.8*5=1960
</span>

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A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the stri

zvonat [6]

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

Here

n = Number of node

T = Tension

$\mu$ = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 281.2Hz}$

Similarly plug in 2 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 562.4Hz}$

Similarly plug in 3 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)$

$\mathbf{f= 843.7Hz}$

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3 years ago

HEEEYYYYY HELLO CAN YOU HELP ME WITH THIS ? PLEASE?

Lubov Fominskaja [6]

Answer:

Here are the names and symbols

H is Hydrogen

Au is Gold

Potassium is K

Mg is Magnesium

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3 years ago

Special relativity can be used to study an object in which frame of reference? A. A frame of reference that has no change in vel

Georgia [21]

Answer:

gangster

Explanation:

sorrynsjajshdnqmanshsha

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3 years ago

Read 2 more answers

How does the increasing mass effect the force of an object in motion?

irina [24]

Answer:

<u>According </u><u>to </u><u>second </u><u>law </u><u>of </u><u>motion</u><u>,</u><u>t</u><u>he acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.</u>

<em>So </em><em>simply</em><em>,</em><em> </em><em>it </em><em>can </em><em>be </em><em>affected </em><em>due </em><em>to </em><em>increasing </em><em>force </em><em>as </em><em>there </em><em>is </em><em>close </em><em>relationship </em><em>between </em><em>momentum.</em>

Explanation:

<em>The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.</em>

<em>I </em><em>hope </em><em>it </em><em>was </em><em>helpful </em><em>for </em><em>you </em><em>:</em><em>)</em>

7 0

1 year ago

An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration

zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Distance = 53m

Time = 5.2s

Unknown:

Acceleration = ?

Solution:

This is a linear motion and we use the right motion equation;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

Insert the parameters and solve;

53 = (0x 5.2) + $\frac{1}{2}$ x a x 5.2

53 = 2.6a

a = $\frac{53}{2.6}$ = 20.4m/s²

6 0

3 years ago