Which of the following displays has the highest hz frequency

A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later

UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]

= (64.2 m/s) - [ 14.7 m/s ]

= 49.5 m/s . (upward)

7 0

2 years ago

A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.

zmey [24]

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

8 0

2 years ago

Read 2 more answers

Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an

blsea [12.9K]

They are both right because you can note both things, I mean Raphael and Lucinda, both has a right statement or explanation about the wave. Wave by nothing is both for its wavelength and for its frequency. So Raphael and Lucinda are both correct because you can note both wavelength and frequency.

7 0

2 years ago

For what value m of the clockwise couple will the horizontal component ax of the pin reaction at a be zero? if a couple of that

sergejj [24]

Thank you for posting your question here at brainly. Below is the answer:

sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M
- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) 

sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M 
- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) 

Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) 
= (2M + 424.91)/(5.745) 

M = ((275*5.745) - 424.91)/2 
= 577.483 which rounds off to 577 

Is it maybe supposed to be Ay - Cy = 275

8 0

2 years ago

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

2 years ago