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OLEGan [10]
3 years ago
9

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true abo

ut them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.
Physics
1 answer:
ycow [4]3 years ago
7 0

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

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Select the correct answer. Which of the following is a category of dance? O A. Group dance OB. Solo dance O C. Partner dance OD.
jeka57 [31]

Answer:

all of above I think sorry if I'm not right but I tried

3 0
3 years ago
A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what
andreyandreev [35.5K]

Answer:

The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.

when angle is zero then flux is maximium because when angle zerocos is maximium

8 0
3 years ago
A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
IrinaVladis [17]

The coefficient of friction between the road and the car's tire is determined as 0.78.

<h3>Acceleration of the car</h3>

The acceleration of the car is calculated as follows;

v² = u² - 2as

0 = u² - 2as

a = u²/2s

where;

  • u is the initial velocity = 97 km/h = 26.94 m/s

a = (26.94)²/(2 x 47)

a = 7.72 m/s²

<h3>Coefficient of friction</h3>

μ = a/g

μ = (7.72)/9.8

μ = 0.78

Learn more about coefficient of friction here: brainly.com/question/14121363

#SPJ1

5 0
2 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie
polet [3.4K]

Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz

8 0
3 years ago
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