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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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Σ/ε
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Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what
kobusy [5.1K]
<span>b) The force with a distance of 150 km is 889 N c) The force with a distance of 50 km is 8000 N This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question. Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as F = (G M1 M2)/r^2 where F = Force G = gravitational constant M1 = Mass 1 M2 = Mass 2 r = distance between center of masses for the two masses. So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889. Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits. If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
8 0
3 years ago
Read 2 more answers
How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec     time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0
1 year ago
In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a
matrenka [14]

Answer:

Approximately \displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right].

Explanation:

Consider this 45^{\circ} slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin (0, 0).

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at 45^{\circ} to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as y = -x.

Convert the initial speed of this diver to SI units:

\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}.

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at g (g \approx \rm -9.81\; m\cdot s^{-2} near the surface of the earth.) At t seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

  • x-coordinate: 30.556t meters (constant velocity;)
  • y-coordinate: \displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2} meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate t from this expression, solve the equation between t and x for t. That is: express t as a function of x.

x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}.

Replace the t in the equation of y with this expression:

\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}.

Plot the two functions:

  • y = -x,
  • \displaystyle y= -0.0052535\;x^{2},

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of y (right-hand side of the equation, the part where y is expressed as a function of x.)

-0.0052535\;x^{2} = -x,

\implies x = 190.35.

The value of y can be found by evaluating either equation at this particular x-value: x = 190.35.

y = -190.35.

The position vector of a point (x, y) on a cartesian plane is \displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]. The coordinates of this skier is approximately (190.35, -190.35). The position vector of this skier will be \displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]. Keep in mind that both numbers in this vectors are in meters.

4 0
3 years ago
Is 0.0 m/s squared increasing or decreasing
Agata [3.3K]
Decrease because its 0.0 m/s
4 0
3 years ago
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