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Ilya [14]
3 years ago
8

A 2.0 kg stone is tied to a 0.30 m string and swung around a circle at a constant angular velocity of 12.0 rad/s. The net torque

on the stone about the center of the circle is?
Physics
1 answer:
saul85 [17]3 years ago
6 0

Answer: τ = 0

Explanation:

At constant angular velocity there is no angular acceleration therefore no torque.

τ = Iα

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Ahat [919]
Lysosomes are used by the cell to digest or breakdown multifaceted organic molecules
8 0
3 years ago
Read 2 more answers
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

4 0
3 years ago
A supertanker ( = 1.70 × 108 kg) is moving with a constant velocity. Its engines generate a forward thrust of 7.40 × 105 N. Dete
a_sh-v [17]

Answer:

(a) 0 (b) F=1.67\times 10^9\ N

Explanation:

Given that,

Mass of a supertanker, m=1.7\times 10^8\ kg

The engine of a generate a forward thrust of, F=7.4\times 10^5\ N

(a) As the supertanker is moving with a constant velocity. We need to find the magnitude of the resistive force exerted on the tanker by the water. It is given by :

F = ma, a is the acceleration

For constant velocity, a = 0

So, F = 0

(b) The magnitude of the upward buoyant force exerted on the tanker by the water is equal to the weight of the ship.

F = mg

F=1.7\times 10^8\times 9.8\\\\F=1.67\times 10^9\ N

Hence, this is the required solution.

4 0
2 years ago
A block of mass m is compressed against a spring (spring constant kk) on a horizontal frictionless surface. The block is then re
Nimfa-mama [501]

Answer:

A) True, B) False, C) False  and  D) false

Explanation:

Let's solve the problem using the law of conservation of energy to know if the statements are true or false

Let's look for mechanical energy

Initial

     Emo = Ke = ½ k Dx2

Final

     Em1= ½ m v12

     Emo = Em1

     ½ k Δx2 = ½ m v₁²

    v₁² = k / m Δx²

    v₁ = √ k/m   Δx

Now let's calculate the speed when it falls

   Vfy² = Voy² - 2gy

   Vfy² = - 2gy

   Vf² = v₁² + vfy²

A) True     v₁ = A Δx

.B) False. As there is no rubbing the mechanical energy conserves

.C) False the velocity is proportional to the square root of the height

     v2y = v2 √2

. D) false promotional compression speed

3 0
3 years ago
A 2000 kg car experiences a constant braking
bezimeni [28]
Solving for acceleration:

F = ma
10000 = 2000*a
a = 5 m/s^2

Solving for velocity:

Vf = Vi + a(t)
0 = Vi - (5)(6)
Vi = 30 m/s

Solving for displacement or distance:

S = Vi*t - 1/2 * a * t^2
S = 30(6) - 1/2 (5)(6)^2
S = 90m

The car would have traveled 90m before coming to rest.
5 0
3 years ago
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