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3 years ago

At positions and_, _kinetic energy is maximum. At position

Physics

2 answers:

goblinko [34]3 years ago

8 0

Answer:

<h2><em>At positions A and G kinetic energy is maximum. At position D, potential energy is maximum.</em></h2>

Explanation:

Kinetic energy can be defined as the energy of a body due to its movement. This energy depends directly on the speed, the more velocity, more kinetic energy will be.

Potential energy is defined as the energy of a body due to its position where the velocity is zero, for example, the position at 12 meters above sea level.

Now, in this case, at points A and G the kinetic energy is maximum because at such moment, the speed of the movement is the highest.

The point D have the maximum potential energy, because there the ball has no velocity, at that moment the ball instantly stops.

So, the complete paragraph is

<em>At positions A and G kinetic energy is maximum. At position D, potential energy is maximum.</em>

worty [1.4K]3 years ago

7 0

Hi there!

I believe it looks like this.

At positions A, G kinetic energy is maximum.
At position D, potential energy is maximum.

Hope this helps! ☺♥

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An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce

ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

From them we can get $v^2=v_0^2+2ad$ .

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}$

We multiply both sides by 2a, and continue:

$2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Being d the displacement $x-x_0$ , we have $v^2=v_0^2+2ad$

For our exercise, we will write this as:

$d=\frac{v^2-v_0^2}{2a}$

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

$d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m$

For the time we use:

$t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s$

6 0

4 years ago

Sound level B in decibels is defined as

Lelechka [254]

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

3 0

3 years ago

what english word does the following- the first two signify a male, the 1st 3 letters support a female, the first four support a

Zarrin [17]

Answer:

lol i know - i was rushing -_-

he= male

her=female

One word with both is <u>heroin</u> but Im not 100% sure

8 0

3 years ago

In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of

Airida [17]

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

3 0

3 years ago

Plsss help me!!!!!!!!!

natali 33 [55]

It should be the B
Low frequency and long wavelength

6 0

3 years ago

At positions and_______, _______kinetic energy is maximum. At position

At positions and_, _kinetic energy is maximum. At position