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Romashka [77]
3 years ago
14

A hedgehog runs 500 meters in 10 minutes, then 300 meters in 8 minutes and finally 100 meters in 6 minutes. What is the hedgehog

's average speed in meters per minutes?
Physics
1 answer:
Marina86 [1]3 years ago
5 0

200 meters per minutes

500 meters -300 meters= 200 meters

300 meters -100 meters =200 meters

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5.6 g/ml. That is the density.
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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What
Crank

Answer:

\mu = \frac{r (2\pi f)^{2}}{g}

Explanation:

N = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

N = mg

f = frequency of rotation

Angular velocity of turntable is hence given as

w = 2\pi f

r = distance from the axis of rotation

\mu = minimum coefficient of static friction

static frictional force is given as

f = \mu N\\f = \mu mg

The  static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}

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3 years ago
If you take a deep breath of air, you can float more easily in water because you have _____.
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Answer:

the answer is c

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A 1-kg rock is suspended by a massless string from one end of a
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Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, m_{r} = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

F_{r} d - F_{s}d = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

F_{s} =F_{r}

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thus, the weight of measuring stick is 9.8 N

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