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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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Answer:

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Explanation:

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5 0
3 years ago
Phase unbalance causes three-phase motors to operate at temperatures ? and, therefore, the motor cannot deliver its rated horsep
anastassius [24]

Phase unbalance causes three-phase motors to operate at temperatures higher than nameplate ratings and, therefore, the motor cannot deliver its rated horsepower.

<h3>What is an electric vehicle (EV)?</h3>

An EV stands for an electric vehicle. Electric vehicles (EVs) are autos that are powered totally or partially by electricity.

Electric cars are extremely cost-effective to run because they have fewer moving parts to maintain and use little to no fossil fuels (petrol or diesel).

Hence, three-phase motors operating with phase imbalance run at temperatures over nameplate limits, which prevents the motor from producing its rated horsepower.

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7 0
2 years ago
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plan
pshichka [43]

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

6 0
3 years ago
Which law of thermodynamics would be violated if heat were to spontaneously flow between two objects of equal temperature?
maks197457 [2]
<span>Actually the second law of thermodynamics would truly gets violated ie, which means that the entrophy changes of the isolated system can never be negative, which covers the above that if heat were to spontaneously flow between any two objects of equal temperature would be fully violated.</span>
5 0
3 years ago
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different
kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.

Solution:

- The standard atmospheric equation is expressed as:

                           P = P_a* ( 1 - βh/T_a)^(g / R*β)

                          (g / R*β) = 32.174 / 1716*0.0035 = 5.252

                            P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

                            P = 1240 lb/ft^2

- The air density method which is expressed as:

                            P = P_a - γ*h

                            P = 2116.2 - 0.07647*14,110

                            P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

                            P = P_a* e^ ( -g*h / R*T_a )

                            P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

                            P = 1270 lb/ft^2

6 0
3 years ago
Read 2 more answers
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