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uranmaximum [27]
2 years ago
10

A bicyclist is initially traveling at 3 m/s. The bicyclist accelerates at 1 m/s2 for 5 seconds.

Physics
1 answer:
leonid [27]2 years ago
7 0
The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

Distance = (3*5) + (1/2*1*5^2)= 15+12.5= 27.5m
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Un atleta de 70 kg de masa que ha efectuado un salto de altura cae una vez que ha
Allushta [10]

Answer:

a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards

Explanation:

The forces on the athlete are

a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,

therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards

b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles

c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.

3 0
2 years ago
Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird
serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr  and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

  1. db = 2dr
  2. L + (L - x) = 2x
  3. 2L - x = 2x
  4. 2L = 3x
  5. x = \frac{2}{3}L

Insert it in x = \frac{2}{3}L

\frac{2}{3}(2.4km) = 1.6km

Now we use formula db = 2dr

  1. db = 2L - x
  2. db = 2(2.4km) - 1.6km
  3. <u>db = 3.2km</u>

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

  1. Vr = L/ Δt ⇒ Δt = \frac{L}{Vr}
  2. \frac{2.4km}{6.8km/hr}
  3. \frac{6}{17}hr

(Km cancel each other)

  1. Vb = db/Δt ⇒ db = VbΔt
  2. 13.6km/hr(\frac{6}{17}hr )
  3. <u>4.8km</u>

(hr cancel each other)

Hope it helps you :)

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Ann [662]

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2 years ago
Please answer this<br> No links please
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Your answer is- amplitudes
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