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2 years ago

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.

Physics

2 answers:

Nesterboy [21]2 years ago

8 0

A = v²/r

v = 3.3 m/s
r = 0.13 m

a: centripetal acceleration

Aleks [24]2 years ago

3 0

By definition, centripetal acceleration is given by:

$a = \frac{v ^ 2}{r}$

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

$a =\frac{3.3^2}{0.13}\\a = 83.76923077$

Rounding the result we have:

$a = 83.8 \frac{m}{s^2}$

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

$a = 83.8 \frac{m}{s^2}$

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riadik2000 [5.3K]

The temperature and the solubility of sugar at that temperature

Explanation:

The amount of substance which can be dissolved in the solvent depends on the temperature.

As the temperature increases, more substance can be dissolved.

A solution is saturated if any more of the solute cannot be dissolved in the solution at the given temperature

Hence we need to know the temperature and also the amount of substance which can be dissolved(solubility) at the same temperature

a) the statement given in option A is correct

b) molar mass has no correlation with the substance's solubility and hence option b is not correct

c) The percent by volume of the solution is not needed to find if the solution is saturated and hence option c is not correct

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2 years ago

550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c

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Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

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Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work

7 0

2 years ago

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

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Starting from the angular velocity, we can calculate the tangential velocity of the stone:
$v=\omega r= (12 rad/s)(0.5 m)= 6 m/s$

Then we can calculate the angular momentum of the stone about the center of the circle, given by
$L=mvr$
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
$L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}$

3 0

2 years ago

A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A seco

Roman55 [17]

Answer:

Explanation:

Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .

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B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)

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= B x area of face of coil x no of turns of secondary

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