A 25 kg cart has 125 kg*m/s of momentum. How fast is the car going?

Suppose you place a ball in the middle of a wagon, and then accelerate the wagon forward. Describe the motion of the ball relati

rjkz [21]

Answer:

The motion of the ball relative to the ground is stationary

The motion of the ball relative to the wagon is backwards

Explanation:

To describe the motion of the ball relative to the ground, we note that

Assuming the ball is perfectly round and rotate freely, then we have

Force on the ball due to motion of the wagon = 0 N,

Then by the law of motion, an object will remain at rest when no force is applied to it

Therefore, apart from rotation of the ball, it will remain no displacement relative to the ground.

The motion of the ball relative to the wagon

Relative to the wagon, the ball appears to be moving in the opposite direction to the wagon, that is backwards.

4 0

3 years ago

Read 2 more answers

A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci

Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0

2 years ago

You perform an experiment with a long column of air and a tuning fork. The column of air is defined by a very long vertical plas

velikii [3]

Answer:

$\lambda=4L=1.33m$

v=343m/s

Explanation:

We have to take into account the expressions

$f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}$

if we assume that 256Hz is the fundamental frequency we have

$f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m$

and for wavelength

$\lambda=4L=1.33m$

hope this helps!!

6 0

3 years ago

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A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$