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zalisa [80]
3 years ago
8

An amusement park ride consists of a rotating circular platform 10.9 m in diameter from which 10 kg seats are suspended at the e

nd of 2.81 m massless chains. When the system rotates, the chains make an angle of 26.5◦ with the vertical.The acceleration of gravity is 9.8 m/s2 .What is the speed of each seat? Answer in units of m/s.(2) If a child of mass 40.8 kg sits in a seat, what is the tension in the chain (for the same angle)?Answer in units of N.
Physics
1 answer:
Darya [45]3 years ago
4 0

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A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve
Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

s_2 = gt^2/2

s_1 = g(t + \Delta t)^2/2

v_2 = gt

v_1 = g(t + \Delta t)

Their difference in distance are therefore:

\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2

\Delta s = g/2((t + \Delta t)^2 - t^2)

\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t) (AsA^2 - B^2 = (A-B)(A+B)

\Delta s = g\Delta t/2(2t + \Delta t)

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

\Delta v = v_1 - v_2 = g(t + \Delta t) - gt

\Delta v = gt + g\Delta t - gt = g\Delta t

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0
3 years ago
Read 2 more answers
A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed
Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0
3 years ago
Io experiences tidal heating primarily because __________. hints io experiences tidal heating primarily because __________. io i
maxonik [38]
Lo experiences tidal heating primarily because lo’s elliptical orbit causes the tidal force on lo to vary as it orbits the Jupiter. Thus, lo’s elliptical orbit is essential to its tidal heating. This elliptical orbit, in turn, is an end result of the orbital resonance among lo, Europa and ganymade. This orbital resonance origin lo to have a more elliptical orbit than it would because lo intermittently passes Europa and ganymade in the same orbital position. We cannot perceive tidal forces of tidal heating in lo but rather we foresee that they must occur based on the orbital characteristic of the moons and active volcanoes on lo is the observational evidence that tidal heating is significant in lo.
8 0
3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

c.Work done by friction on the incline,w_{friction}=P.E-spring \;energy

W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has be
ale4655 [162]

Answer:

500J

Explanation:

The arrow will have an energy of 500J after it has been released from its state of rest.

This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".

  • The energy at rest which is the potential energy is 500J
  • This energy will be converted to kinetic energy in total after the arrow has been released.
  • This way, no energy is lost and we can account for the energy transformations occurring.
4 0
3 years ago
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