A car that increases it's speed from 20 km/h to 100 km/h undergoes

What is your acceleration near earth due to gravity

Maksim231197 [3]

I dont know but i know i dont lnow if this is true but the gravity is slowly going away every 4 year i dont know i think.

7 0

2 years ago

What force is left out of the Quantum Mechanics theory?

nirvana33 [79]

Answer:

Quantum mechanics is a key hypothesis in material science that gives a portrayal of the actual properties of nature at the size of iotas and subatomic particles. It is the establishment of all quantum physical science including quantum science, quantum field hypothesis, quantum innovation, and quantum data science.

Explanation:

It is the greatest of issues, it is the littlest of issues. At present physicists have two separate rule books clarifying how nature functions. There is general relativity, which perfectly represents gravity and everything it overwhelms: circling planets, impacting worlds, the elements of the growing universe all in all. That is enormous. At that point there is quantum mechanics, which handles the other three powers – electromagnetism and the two atomic powers. Quantum hypothesis is very proficient at portraying what happens when a uranium molecule rots, or when singular particles of light hit a sun based cell. That is little.

4 0

2 years ago

I am boring and you can talk to me?

sladkih [1.3K]

SURE HI

HI HOW R U

Hello

4 0

2 years ago

Read 2 more answers

A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th

finlep [7]

Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

$\omega _{0}=\frac{1}{\sqrt{LC}}$

$\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}$

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = $\sqrt{R^{2}+\left ( XL - Xc \right )^{2}}$

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

7 0

2 years ago

A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0

otez555 [7]

A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
$x(t)=v_0 \cos \alpha t$
$y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
where
- the horizontal motion is a uniform motion, with constant speed $v_0 \cos \alpha$ , where $v_0 = 8.00 m/s$ and $\alpha=20.0^{\circ}$
- the vertical motion is an uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ , initial position h (the height of the building) and initial vertical velocity $v_0 \sin \alpha$ (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
$x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m$

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
$0=h-v_0 \sin \alpha t - \frac{1}{2}gt^2$
which becomes
$h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m$

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
$y(t)=h-10$
If we substitute this into the equation of y(t), we have
$h-10 = h-v_0 \sin \alpha t- \frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0$
$4.9 t^2 +2.74 t-10 =0$
whose solution is $t=1.18 s$ (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0

3 years ago