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2 years ago

Can someone please help me get this right pleaseee I’ll mark brainless .

Physics

2 answers:

Katarina [22]2 years ago

4 0

Answer:i think it is c

Explanation:

Papessa [141]2 years ago

4 0

Answer:

Explanation: i think its c to try it

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The graph illustrates the activity level of three common digestive enzymes, across a range of pH values. Which enzyme is likely

vaieri [72.5K]

Answer:

(A) Pepsin

Explanation:

From the graph it is clear that pepsin is the only enzyme which works in highly acidic condintion in the digestive system.

less than 7 the liquid is acidic
above 7 the liquid is basic
at 7 the liquid is neutral

It has an optimum pH of about 1.5 at which its activity level is 8.5 as shown in graph.

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3 years ago

Which of the following was NOT one of the top 10 reported Stressors for Teens?

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3 years ago

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago

The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a

Setler [38]

Answer:

x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

$x_{cm}$ = 1 / $M_{total}$ ∑ $x_{i} m_{i}$

Where M_{total} is the total masses of the system, $x_{i}$ is the distance between the particles and $m_{i}$ is the masses of each body

Let's apply this equation to our problem

M = Me + m

M = 5.98 10²⁴ + 7.36 10²²

M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]

x_{cm} = 4.644 10⁶ m

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2 years ago

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Fofino [41]

Explanation:

Below is an attachment containing the solution.

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2 years ago