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Fiesta28 [93]
3 years ago
11

A heavy dart and a light dart are launched vertically by identical ideal springs. Both springs were initially compressed by the

same amount. There is no significant air resistance. Which of the following statements about these darts are correct? (There could be more than one correct choice.)a. Both darts reach the same maximum height. b. The heavy dart goes higher than the light dart. c. Both darts began moving upward with the same initial speed. d. The light dart goes higher than the heavy dart At the maximum height, both darts have the same gravitational potential energy.
Physics
1 answer:
sveticcg [70]3 years ago
8 0

Answer:

D

Explanation:

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A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and
Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

         W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

         Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         mg h = K

to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

            h = L (1-cos 35)

we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

7 0
3 years ago
Carlos gives a grocery cart a 60-N push. The cart has a mass of 40 kg. What is the cart's acceleration?
andre [41]
<h3>Answer:</h3>

1.5 m/s²

<h3>Explanation:</h3>

We are given;

Force as 60 N

Mass of the Cart as 40 kg

We are required to calculate the acceleration of the cart.

  • From the newton's second law of motion, the rate of change in momentum is directly proportional to the resultant force.
  • That is, F = ma , where m is the mass and a is the acceleration

Rearranging the formula we can calculate acceleration, a

a = F ÷ m

  = 60 N ÷ 40 kg

  = 1.5 m/s²

Therefore, the acceleration of the cart is 1.5 m/s²

3 0
3 years ago
Shanika is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. I
Whitepunk [10]

Answer:

I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy

Explanation:

8 0
3 years ago
Physic help????????????????
maxonik [38]

My personal understanding and opinion is that ALL of those questions
should be part of an assessment of Physical Activity Readiness.


4 0
3 years ago
Read 2 more answers
A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

4 0
3 years ago
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