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vekshin1
3 years ago
11

Velocity and Time

Physics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer:

h=112.35

Explanation:

becuase it had to

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Identify the breed of goat in the picture.
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Answer:

look it up im not a sheaperd sorry

Explanation:

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2 years ago
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a 2000 kg truck is traveling at a velocity of 30 m/s. What velocity must a 1000 kg car have in order to have the same momentum a
Molodets [167]

The car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

Answer:

Explanation:

Momentum is measured as the product of mass of object with the velocity attained by that object.

Momentum of 2000 kg truck = Mass × Velocity

Momentum of 2000 kg truck = 2000×30 = 60000 N

Similarly, the momentum of 1000 kg car will be 1000× velocity of the 1000 kg car.

Since, it is stated that momentum of 2000 kg truck is equal to the momentum of 1000 kg of car, then the velocity of 1000 kg of car can be determined by equating the momentum of car and truck.

Momentum of 2000 kg truck = Momentum of 1000 kg car

60000=1000×velocity of 1000 kg car

Velocity of 1000 kg car = 60000/1000=60 m/s

So, the car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

4 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

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3 years ago
Is a reflection matter?
azamat

Answer:

Yes, because everything bounce off in every surface around any object.

Explanation:

7 0
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