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Arte-miy333 [17]
4 years ago
15

If the normal force exerted by the track on the car when it is at the top of the track (point BB) is 6.00 NN , what is the norma

l force on the car when it is at the bottom of the track (point AA)?
Physics
1 answer:
Setler79 [48]4 years ago
4 0

Answer:

N_A=21.68N

Explanation:

We already know that the wight in reference is <em>gravitational</em> force on an object:F_G=mg.

Gravity=9.8m/s^2

Weight of car=0.800kg

To solve for A, we apply Newton's second law of radial direction:

\sum F_r_a_d=ma_R\\=mg+N_B\\

Rewrite to calculate acceleration,a:

a_R=\frac {mg+N_B}{m}\\=\frac{0.8kg+9.8m/s^2+6N}{0.8kg}=17.3m/s^2

Solving for Position A:

\sum F_r_a_d=ma_R\\=N_A-mg\\N_A=ma_R+mg=0.8kg(17.3m/s^2+9.8m/s^2)\\=21.62N

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A wave travels along a stretched horizontal rope. The vertical distance from crest to trough for this wave is 13 cm and horizont
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Answer:

(A) The wavelength of this wave is 56\; \rm cm.

(B) The amplitude of this wave is 6.5\; \rm cm.

Explanation:

Refer to the diagram attached. A point on this wave is at a crest or a trough if its distance from the equilibrium position is at a maximum.

The amplitude of a wave is the maximum displacement of each point from the equilibrium position. That's the same as the vertical distance between the crest (or the trough) and the equilibrium position.

  • On the diagram, the distance between the two gray dashed lines is the vertical distance between a crest and a trough. According to the question, that distance is \rm 13\; \rm cm for the wave in this rope.
  • On the other hand, the distance between either gray dashed line and the black dashed line is the distance between a crest (or a trough) and the equilibrium position. That's the amplitude of this wave.

Therefore, the amplitude of the wave is exactly \displaystyle \frac{1}{2} the vertical distance between a crest and a trough. Hence, for the wave in this question,

\begin{aligned}& \text{Amplitude}\\ &= \frac{1}{2} \times (\text{Vertical distance between crest and trough}) \\ &= \frac{1}{2} \times 13\;\rm cm = 6.5\; \rm cm\end{aligned}.

The wavelength of a transverse wave is the same as the minimum (horizontal) distance between two crests or two troughs. That's twice the horizontal distance between a crest and a trough in the same period.

\begin{aligned}& \text{Wavelength}\\ &= 2 \times (\text{Horizontal distance between adjacent crest and trough}) \\ &= 2 \times 28\;\rm cm = 56\; \rm cm\end{aligned}.

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