If a horizontal force of 15N applied to a 50N block towards the east is just enough to cause that block to move

A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system applicat

umka21 [38]

Answer:

$Av_2 =24\ V/V$

Explanation:

given,

op-amp circuit with a gain of = (Av₁) = 96 V/V

Band width = (Bw₁) = 8 kHz

Required bandwidth(Bw₂) = 32 kHz

Highest gain available =(Av₂) = ?

For the given system Bandwidth product is constant

Av₁ Bw₁ = Av₂ Bw₂

96 x 8 = Av₂ x 32

$Av_2= \dfrac{96\times 8}{32}$

$Av_2 =24\ V/V$

the highest gain available under these conditions $Av_2 =24\ V/V$

6 0

3 years ago

How many pounds of meat does a cougar eat per day

Luba_88 [7]

A large male cougar living in the Cascade Mountains kills a deer or elk every 9 to 12 days, eating up to 20 pounds at a time and burying the rest for later.Except for females with young, cougars are lone hunters that wander between places frequented by their prey, covering as much as 15 miles in a single night.Cougars rely on short bursts of speed to ambush their prey. A cougar may stalk an animal for an hour or more

hope this helps in any way ! :)

4 0

3 years ago

Which statement is the correct representation of these electric field lines?

GuDViN [60]

I would say a. hope this helps

4 0

3 years ago

Read 2 more answers

The wavelengths for visible light rays correspond to which of these options?

ryzh [129]

Answer:

sorry about the other person but its b

Explanation:

5 0

2 years ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

2 years ago