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WITCHER [35]
3 years ago
11

When a car drives over a speed bump and oscillates up and down in simple harmonic motion, at which position during the motion is

the acceleration of the car the greatest? when a car drives over a speed bump and oscillates up and down in simple harmonic motion, at which position during the motion is the acceleration of the car the greatest? at the equilibrium position, x = 0 at half the maximum amplitude, x = a/2 at the maximum amplitude, x = a none; the acceleration is constant?
Physics
1 answer:
Oksana_A [137]3 years ago
3 0
Using the second Law of Newton, F = m * a, you know that acceleration is maximum when the force is maximum.


Using Hooke's Law, F = K Δx, you know that the force is maximum when the displacement from the equilibrium (Δx) is maximum.


So the answer is that the acceleration is maximum at the maximum amplitude x = a.
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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
3 years ago
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

  • m is mass of the rocket
  • v is the final velocity of the rocket
  • u is the initial velocity of the rocket
  • t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

7 0
2 years ago
Question 9 (28 points)
ladessa [460]
Im sure the answer is letter B
5 0
3 years ago
Read 2 more answers
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0
3 years ago
Read 2 more answers
What causes dispersion of light?​
AysviL [449]
Dispersion occurs due to the different degrees of refraction experienced by different colours of light. Light of different colours may travel with the same speed in a vacuum, but they travel at different speeds in some refracting medium. The speed of violet light is relatively lower than that of red light.
7 0
3 years ago
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