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vladimir1956 [14]
2 years ago
12

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Physics
1 answer:
harina [27]2 years ago
6 0

Answer:

4.437 m/s

Explanation:

Diameter of rotation d is 1.7 m

Radius of rotation = d/2 = 1.7/2 = 0.85 m

If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s

We convert to rad/s

Angular speed = 2 x pi x 0.83

= 2 x 3.142 x 0.83 = 5.22 rad/s

Speed is equal to the angular speed times the radius of rotation

Speed = 5.22 x 0.85 = 4.437 m/s

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Answer it pls!!!!!!!!!!!
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Answer:

Fractional error = 0.17

Percent error = 17%

F = 112 ± 19 N

Explanation:

Plug in the values to find the force:

F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N

Find the fractional error:

ΔF/F = Δm/m + 2Δv/v + Δr/r

ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5

ΔF/F = 0.17

Multiply by 100% to find the percent error:

ΔF/F × 100% = 17%

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ΔF = 0.17 × 112 N = 19 N

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Answer:

Is there a map of the town? It doesn't make sense without coridnates or street lengths. If it was a straight isosilese triangles the return time is 3:36pm. This answer doesn't make sense if there is a map.

Explanation:

Side a is 12km Side b 15 km Side c 19.21 km.

9am depearture

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2 years ago
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
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Answer:

A) 26.5 m/s

B) 33.0 m/s

Explanation:

A)

  • Once the car leaves the cliff, as no other influence than gravity acts on it, and since it causes the car an acceleration in the vertical direction only, in the horizontal direction, it keeps moving at the same speed until it reaches to the other side.
  • So, we can apply the definition of average velocity to find this speed as follows:

       v_{x} = \frac{\Delta x}{\Delta t}  (1)

  • We know the value of Δx, which is just the wide of the river (53.0m), but we need to find also the value of Δt.
  • This time is given by the vertical movement, whic.h is independent from the horizontal one, because both movements are perpendicular each other.
  • Since the only influence in the vertical direction is due to gravity, the car is accelerated by gravity, with constant acceleration downward equal to g = -9.8m/s² (taking the upward direction as positive).
  • Since the acceleration is constant, we can use the following kinematic equation, as follows:

       \Delta y  = y_{f} - y_{o} = v_{o} * t + \frac{1}{2}  * g *t^{2}  (2)

  • if we take the river level as our x-axis, this means that yf = 1.3 m and

       y₀ = 20.8 m.

  • At the same time, due to in the vertical direction the car has no initial velocity, this means that  v₀ = 0.
  • Replacing by the values in (2) , and solving for t:

       t = \sqrt{\frac{2* \Delta y}{g} } = \sqrt{\frac{2*19.5m}{9.8m/s2} }  = 2 s  (3)

  • If we choose t₀ =0 ⇒ Δt = t = 2 s
  • Replacing Δx and Δt in (1):

       v_{x} = \frac{\Delta x}{\Delta t} = \frac{53.0m}{2s} = 26.5 m/s  (4)

B)

  • When the car is just landing in the other side, the velocity of the car has two components, the horizontal one that we just found in A) and a vertical one.
  • Due to the initial velocity in the vertical direction was just zero, we can find the final velocity just applying the definition of acceleration, with a =g, as follows:

      v_{fy} = g*t = -9.8m/s2*2 s = -19.6 m/s  (5)

  • Since both components are perpendicular each other, we can find the magnitude of the velocity vector (the speed) using the Pythagorean Theorem, as follows:

       v = \sqrt{v_{x}^{2} + v_{fy}^{2} } } = \sqrt{(26.5m/s)^{2} + (-19.6m/s)^{2}}  = 33.0 m/s  (6)

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Answer:

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Facilitated diffusion is the passive (no energy required) movement of molecules across a cell membrane, with the aid of a membrane protein. energy in form of ATP or GTP is not required since molecules are moving down their concentration gradient.

The need for facilitated diffusion occurs when substances cannot successfully cross the phospholipid bilayer of the cells membrane on their own either due to their polarity or large size. The proteins in facilitated diffusion are tthose that act as carriers and those the form the channels across the membrane, through which the diffusion occurs. Examples of where facilitated diffusion is seen is in the transport of glucose (relatively large molecule) across the cell membranes, with the help of the glucose transporter and the transportation of water molecules across the cell membrane through aquaporins.

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