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vova2212 [387]
3 years ago
8

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

e and brings it up to its proper turning rate of one complete rotation every 6.3 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding
Physics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

u_{s}=0.56

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

F_{s.max}=\frac{mv^2}{r} \\

Where the r is the radius of merry-go-round and v is the tangential speed

but

F_{s.max}=u_{s}F_{N}=u_{s}mg

So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

Substitute the given values

So

u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56

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When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g
Olegator [25]

Answer:

Your question was incomplete so here is the complete question and answer.

Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)

a) plain water

b) 5-7 percent glucose solution

c) Glucose polymer solution of 6-8 percent

d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Explanation:

Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.

3 0
3 years ago
The gas used to fill party ballons contain only helium atoms .This make helium a(n).
Nata [24]

a gas

hope this is right

5 0
3 years ago
Read 2 more answers
Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m
ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

 a = 62.5²/(2 × 22)

 a = 88.78m/s²

the time you need to get this speed

     v = v₀ + a t

     t = v / a

     t = 62.5 / 88.78

     t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

     t = 1,025 s

      a = 55.43 m / s²

4 0
3 years ago
A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.
Maksim231197 [3]

Answer:

v=33.764m/s

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

F_{c}=\frac{mv^{2} }{r}

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s

4 0
3 years ago
The horizontal surface on which the objects slide is frictionless. If M = 1.0 kg and the magnitude of the force of the small blo
sweet-ann [11.9K]

Answer:

The force is 7.8 N.

Explanation:

Given that,

Mass of small object = 2M

Mass of large object = 4M

Here, M = 1.0 kg

Force of the small block = 5.2 N

We need to calculate the acceleration of 4 kg block

Using formula of force

F=ma

a=\dfrac{F}{m}

a=\dfrac{5.2}{4}

a=1.3\ m/s^2

The 2 kg block is also accelerating at 1.3 m/s², making a total of 6 kg.

We need to calculate the force

Using formula of force

F=ma

Put the value into the formula

F=6\times1.3

F=7.8\ N

Hence, The force is 7.8 N.

5 0
3 years ago
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