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DanielleElmas [232]
3 years ago
5

I NEED HELP ASAP WILL MARK BRAINLIEST:

Physics
1 answer:
Evgen [1.6K]3 years ago
7 0

Answer:

It is b for sure.

Explanation:

because they are examining urine now that's pure science.

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One problem with digital data is that it can be vulnerable to hackers.
hoa [83]

Answer: c. A person who gains unauthorized access to digital data

Explanation:

5 0
1 year ago
You exert a force of 75 newtons on a rock. You push and you push, but you can’t budge it. You are exhausted! How much work did y
dalvyx [7]

Answer:   Work W = 0

Explanation: Work W = F·s. Because rock does not move, s = 0 and

work done is zero.

4 0
2 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
2 years ago
A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the
lilavasa [31]

Answer:

R=803k\Omega

Explanation:

We have the following information,

V_0 = 12V\\V=10V\\c= 1.25*10^{-6}F\\t=1.8s

We apply the equation for capacitor charging the voltage across it,

V=V_0 (1-e^{-t/x})\\e^{-t/x}=1-(\frac{V}{V_0})\\-\frac{t}{Rc}=ln(\frac{V}{V_0})\\R=-\frac{t}{ln(\frac{V}{V_0})*c}

Replacing values,

R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}

R=803k\Omega

3 0
3 years ago
The (blank) of vibration of a wave is defined as that which has the lowest frequency
saveliy_v [14]
Fundamental frequency
6 0
3 years ago
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