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Kruka [31]
3 years ago
13

PLEASE NEED HELP What is the net force acting on the race car in the picture: Question 1 options: 10 N to the right 3 N to the l

eft 17 N 3 N to the right

Physics
1 answer:
Nezavi [6.7K]3 years ago
7 0

Use Net Force Equation.

Fnet = F1 + F2 + F3 + F4

In which the following main forces in play are:

F1 = Gravitational Force

F2 = Normal Force

F3 = Thrust Force

F4 = Air resistance

Since this is assumed to happen on a flat surface, the gravitational and normal forces cancel out. So we are left with:

Fnet = F(thrust) + F(air resistance)

Now we can use this equation to solve for the net force.

Fnet = 10N [right] + 7N [left]

Fnet = 10N [right] - 7N [right]

Fnet = 3N [right].

Therefore, the total net force acting on this car would be 3N [right]

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Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when
nignag [31]
<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s

So, car X will move with a velocity of -12 m/s.

3 0
3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











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Explanation:

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v^2-u^2=2as

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Answer: The answer is D

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