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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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mihalych1998 [28]
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
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3 years ago
Just asking how everyone is doing cause of the virus
Bess [88]

Answer:

We are all fiine

Explanation:

How are you doing also

3 0
3 years ago
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A train travels 250 km westward from Carthage to Johnson City. The train arrives 2.5 hours after it left. What was the average v
den301095 [7]
The answer is C 100 km/h west
5 0
3 years ago
Frame S' passes frame S in the usual way (positive directions). An object also moves in the positive direction. Which is true? T
lisov135 [29]

Answer:

The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."

Explanation:

The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame

take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2

relative velocity of the object to the S' frame would be

Vrel = v2- v

This means the Vrel of the object with respect to the S' frame is less than the Vrel  of the object with respect to the S frame

However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.

This would mean the second option is the answer, the relative speed of the object depends on the actual values.

3 0
3 years ago
Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc
TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0
3 years ago
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