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melisa1 [442]
3 years ago
12

6. A car is traveling at 50m/s when it begins to slow down to

Physics
1 answer:
ki77a [65]3 years ago
8 0

Answer:

-6 m/s²

Explanation:

Given:

v₀ = 50 m/s

v = 20 m/s

t = 5 s

Find: a

v = at + v₀

20 m/s = a (5 s) + 50 m/s

a = -6 m/s²

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Two spherical objects have masses of 3.1 x 10^5 kg and 6.5 x 10^3 kg. The gravitational attraction between them is 65 N. How far
nata0808 [166]

Answer:

4.55 x 10⁹m

Explanation:

Given parameters:

Mass of object 1  = 3.1 x 10⁵kg

Mass of object 2 = 6.5 x 10³kg

Gravitational force  = 65N

Unknown:

Distance between them  = ?

Solution:

To solve this problem, we use the expression below from the universal gravitational law;

    Fg  =    \frac{G mass 1 x mass 2}{distance ^{2} }  

   G = 6.67 x 10⁻¹¹

        65  = \frac{6.67 x 10^{11} x 3.1 x 10^{5} x 6.5 x 10^{3}   }{distance^{2} }    

   Distance  = 4.55 x 10⁹m

         

3 0
3 years ago
(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e
Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>
  • Charge 1 (q₁) = +3 C
  • Charge 2 (q₂) = +5 C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 7 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>
  • Electron = 223 electrons
  • Proton = 218 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

But

1 electron = 1.6×10⁻¹⁹ C

Thus,

Net Charge = -5 × 1.6×10⁻¹⁹ C

Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>
  • Charge 1 (q₁) = 2×10⁻⁴ C
  • Charge 2 (q₂) = 8×10⁻⁴ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.3 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

  • Electron = 4 electrons
  • Proton = 4 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

Learn more about Coulomb's law:

brainly.com/question/506926

#SPJ1

6 0
2 years ago
Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .
Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

P_{av}=V_{rms}I_{rms}Cos\phi

As given in the question

CosФ = R / Z

And we know that

V_{rms}=I_{rms}\times Z

So

P_{av}=I_{rms}\times Z\times I_{rms}\times \frac{R}{Z}

P_{av}=I_{rms}^{2}\times R

6 0
3 years ago
After the same car leaves the platform, gravity causes it to accelerate downward at a rate of 9.8 m/s2. What is the gravitationa
Simora [160]
The gravitational force on the car is

(9.8 m/s^2) x (the car's mass in kg).

The unit is newtons.
7 0
3 years ago
If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr
Agata [3.3K]

Answer:

Resistance in the flash tube, R=3.97\times 10^{-3}\ \Omega

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, C=503\ \mu F=503\times 10^{-6}\ F

The time constant of RC circuit is given by :

\tau=RC

R is the resistance in the flash tube

R=\dfrac{\tau}{C}..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

v=\dfrac{d}{\tau}

\tau=\dfrac{d}{v}

\tau=\dfrac{0.001}{500}

\tau=0.000002\ s

Equation (1) becomes :

R=\dfrac{0.000002}{503\times 10^{-6}}

R=3.97\times 10^{-3}\ \Omega

So, the resistance in the flash tube is 3.97\times 10^{-3}\ \Omega. Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
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