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patriot [66]
3 years ago
11

If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?

Physics
1 answer:
Alisiya [41]3 years ago
4 0

Answer:

The spring constant is 3750 N/m  

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

=>

(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m  

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Citrus2011 [14]

Sally's average speed is <u>35.3 mi/h.</u>

Average speed of a body is the total distance traveled in the given time interval.

Express the distance  d traveled in miles.

d=\frac{97 km}{1.6 km/m\\i} \\ = 60 mi

Express the time t traveled in hours.

t  =\frac{102 min}{60 min/hr}\\ = 1.7 h

Calculate the average speed v.

v =\frac{d}{t} \\ =\frac{60 mi}{1.7h} \\   = 35.3 mi/h

Her average speed is 35.3 mi/h, which is less than the speed limit of 65 mi/h.

However, the average speed of an object is different from its instantaneous speed. It could be possible that at the time when the officer apprehended her, Sally could have been travelling at a speed greater than the prescribed speed limit, which would have prompted the officer to issue a speeding ticket to her.

Thus, the average speed of a person cannot be considered as a bench mark for speeding offences, since her instantaneous speed could have been higher than the speeding limit and yet she could have had an average speed less than the speeding limit.


3 0
3 years ago
Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin
Vera_Pavlovna [14]

Answer:

θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

Explanation:

This is an angular kinematic exercise the equation for the angular position

the particle A

       θ = θ₀ + ω₀ t + ½ α t²

They say for the particle B

     w₀B = ½ w₀

     αB = 2 α

In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial

      t´ = t - t_1

l

et's write the equation of particle B

      θ = θ₀ + w₀B t´ + ½ αB t´2

replace

     θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²

     θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²

4 0
3 years ago
What is the magnitude of the electric field on a +2 C charge if it experiences an electric force of 6 N?
MrMuchimi

Answer:

call me 7164013838 hurry bye

6 0
2 years ago
The following three questions refer to a situation in which a driver is in a car that crashes into a solid wall. The car comes t
Andreas93 [3]

Answer:

1) p₀ = 45000 N / s ,   p₀ '= 1800 , b)  I = -45000 N s ,  I = 1800 Ns

Explanation:

Impulse equals the change in momentum

         I = Δp

1) the initial moment of the car

         p₀ = M v

          p₀ = 1500 30

          p₀ = 45000 N / s

the change at the moment is

          Δp = 45000

because the end the car is stopped

moment of the person

          P₀ ’= m v

         p₀ '= 60 30

          p₀ '= 1800

          D₀ '= 1800

2) of the momentum change impulse ratio

        car

              I =  Δp

              I = -45000 N s

        person

              I = Δpo '

              I = 1800 Ns

3) the object that give the momentum to stop the wall motoring

The person is stopped by the impulse given by the car

a) This area is the one that absorbs most of the vehicle impulse

be) If using a safety painter, the time during which the greater force will act, therefore the lessons decrease

c) The air bag helps reduction in the speed of the person relatively quickly.

6 0
3 years ago
Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be
Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

v_1=\dfrac{72.42 }{2.5}=28.96\ km/h

Average speed of Prav,

v_2=\dfrac{119}{1.25}=95.2\ km/h

Difference in average speed of Ajoba and Prav is :

v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0
3 years ago
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