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klemol [59]
3 years ago
5

Suppose you average 75 MPH over the first halfof a drive, and your average speed for the entiredrive is 60 MPH. What was your av

erage speedfor the second half of the drive?
Physics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

x = 45 MPH

Explanation:

given,

Average speed of the first half = 75 MPH

Average speed of entire ride = 60 MPH

Average speed of the second half = ?

let the average speed of the second half = x MPH

now,

average of entire ride is given as 60 mph so,

 \dfrac{75+x}{2} = 60

 75+x= 2\times 60

       75 + x = 120

         x = 120 -75

        x = 45 MPH

hence, the average speed of the second half comes out to be 45 MPH.

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A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse
Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒    Ii*ωi = If*ωf   <em>(I)</em>

where    

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation <em>(I) </em>we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒  m = 35.98 Kg ≈ 36 Kg

5 0
3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

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3 years ago
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How do you calculate the total resistance of a series circuit
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Add all the resistances across the circuit together the calculate the total resistance
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uppose two train cars are moving toward one another, the first with a mass of 150,000 kg and a velocity of 0.300 m/s; the second
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The value was determined to be 0.122 m/s. The velocity of a body or object determines its direction of motion. Speed is a scalar quantity in its most fundamental form.

Velocity is essentially a vector quantity. It is the rate of change in distance. The initial speed of the first train, which has a mass of 150,000 kg, is 0.3 m/s. The second train has an initial speed of -0.120 m/s and a mass of 110,000 kg.

Let v represent the post-collision speed of the connected mass.

Utilize the idea of momentum.

The speed of the trains is constant both before and after a collision.

150.000 + 110.000v 45.000 - 13200 = 260.000 v 31800 = 260.000 v v = 0.122 m/s 150000 x 0.3 - 110000 x 0.120

After colliding, they move at a speed of 0.122 m/s towards the direction of the right.

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Answer:

A

Explanation:

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