. Increasing the pressure on a gas __________ the volume the gas occupies. (Points : 1)

Um relógio de ponteiros funciona durante um mês. Qual é, em N.C, o número de voltas, aproximadamente, que o ponteiro dos minutos

Margaret [11]

I can’t understand I’m sorry but if you need help with math I can help you

7 0

3 years ago

The displacement of a 500 g mass, undergoing simple harmonic motion, is defined by the function :

Delicious77 [7]

The maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.

Simple Harmonic Motion

The given equation of the simple harmonic motion is

$x=3.5 sin (\frac{\pi }{2t} + \frac{5\pi }{4} )$

Data;

ω = π/2

k = 1.254N/m

Solving this

$\frac{dx}{dt} = -3.5 X \frac{\pi }{2} cos (\frac{x\pi t}{2}+\frac{5\pi }{4} )$

Let's calculate the maximum velocity.

$V_{m} =\frac{3.5\pi }{2}$

This is only possible when cos θ = -1

The maximum kinetic energy is

$K_m =\frac{1}{2} mv^2 = \frac{1}{2} X \frac{500}{1000} X \frac{7^2\pi ^2}^{4} ^2$

$w^2 = \frac{k}{m} \\k = w^2m\\k = \frac{\pi ^2}{4} X \frac{500}{1000} \\k =1.254 N/m$

Using the value of spring constant, we can find the maximum potential energy.

$P.E =\frac{1}{2} k x^2\\P.E =\frac{1}{2} X 1.234 X 3.5^2 \\P.E = 7.56 J$

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

ME = K.E + P.E

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

Learn more on simple harmonic motion here;

brainly.com/question/15556430

#SPJ1

8 0

2 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu

Aloiza [94]

In other words a infinitesimal segment dV caries the charge
dQ = ρ dV 

Let dV be a spherical shell between between r and (r + dr): 
dV = (4π/3)·( (r + dr)² - r³ ) 
= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) 
= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) 
drop higher order terms 
= 4·π·r²·dr 

To get total charge integrate over the whole volume of your object, i.e. 
from ri to ra: 
Q = ∫ dQ = ∫ ρ dV 
= ∫ri→ra { (b/r)·4·π·r² } dr 
= ∫ri→ra { 4·π·b·r } dr 
= 2·π·b·( ra² - ri² ) 

With given parameters: 
Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) 
= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) 
= 3.77×10⁻⁸C 
= 37.7nC

6 0

3 years ago

Ball A was carried to the top of a hill in a straight line, while ball B was carried in a longer, zigzag path. At the top of the

marissa [1.9K]

Answer: equal

Explanation:

Don’t know why honestly

5 0

3 years ago

Read 2 more answers

What is helpful in calculating standard deviation?

Oksana_A [137]