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3 years ago

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A 50 kg bear climbed on the tree branch 10 meters above the ground. If the bear descends to 5 meters above the ground, its poten

tial energy will be decreased by
A/ 4900 J
B/ 2450 J
C/ 2540 J
D/ 2500 J

1 answer:

Simora [160]3 years ago

6 0

<h3>Option A. 4900 J or joules</h3>

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Ksivusya [100]

5.77 × $10^1^4$ Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

c = Speed of light = 3 × $10^8$ m/s

λ = Wavelength of light

∴ f = c / λ

f = $\frac{3*10^8}{520 * 10^-^9}$

= 5.77 × $10^1^4$ Hz

Therefore, 5.77 × $10^1^4$ Hz is the green photon's frequency .

Learn more about wavelength here:

brainly.com/question/10728818

#SPJ1

4 0

1 year ago

A 4-kg object is moving with a speed of 5 m/s at a height of 2 m. The kinetic

tatyana61 [14]

Hello!

$\large\boxed{KE = 50J}$

Use the formula for kinetic energy:

$KE = \frac{1}{2}mv^{2}$

Plug in the given mass and velocity:

$KE = \frac{1}{2} (4)5^{2}$

Simplify:

$KE = \frac{1}{2} (100)\\\\KE = 50 J$

7 0

3 years ago

How does the frequency of sound relate to its pitch?

madreJ [45]

Good afternoon!
the answer to that particular question is this
rule
a particular pitch directly corresponds to frequency in that if you have a pitch you will have a high frequency
if you a low frequency you will have a low pitch
both are intertwined in marriage!

6 0

3 years ago

How long would it take for a sound impulse to travel through a copper Rod 20km

Nana76 [90]

Not sure.can you give me a clue?

8 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago