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zvonat [6]
3 years ago
15

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

ss of 120 kg. You are in a bumper car, sitting still. I run into you, and we collide elastically. I end up with a velocity of 2 m/s going in the negative x-direction. What is your mass, and what is your final velocity?
Physics
1 answer:
AlekseyPX3 years ago
5 0

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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Answer:

Hay diversas leyes que podemos usar acá.

Acá sabemos que la vejiga aumenta su tamaño al reducir la presión, esto tiene sentido, pues al haber menos presión, hay menos fuerza que comprime la vejiga, lo que le permite aumentar su volumen.

Acá tenemos una relación inversa de la forma: V = K/P

Una relación inversa donde la presión esta en el denominador y K es un termino que no depende ni del volumen ni de la presión.

Entonces, a medida que aumenta P, el denominador aumenta, por lo que el valor del volumen decrece.

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V = nRT/P

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7 0
2 years ago
A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above
gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

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When the student pulls on the block with force F_p at an angle \theta, the normal force on the block becomes

N  = Mg- F_psin(\theta)

and hence the frictional force becomes

F_s = \mu (Mg- F_psin(\theta)).

Now, as we increase \theta, sin(\theta) increases which as a result decreases the normal force Mg- F_psin(\theta), which also means the frictional force decreases; Hence choice B stands true.

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6 0
3 years ago
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Advocard [28]

Answer:

v=697.2km/h

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

V=\frac{d}{t}

The distance is clearly 1743 km and the time is:

t=2h+30min*\frac{1h}{60min} =2.5h

Thus, the velocity turns out:

v=\frac{1743km}{2.5h}\\ \\v=697.2km/h

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Best regards.

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3 years ago
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Answer:

d.Energy as heat transferred into an object is determined by the amount of work done on the object.

Explanation:

4 0
3 years ago
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A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

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Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

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b. d2=3.6 mm

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c: outer radius

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3 0
3 years ago
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