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natta225 [31]
2 years ago
12

A bicycle tire of a radius of 0.44 m has a piece of gum stuck on its rim. What is the angle through which the tire rotates when

the gum has moved through a linear distance of 1.87 m? Express your answer in radians and degrees.
Physics
1 answer:
Karolina [17]2 years ago
6 0

Based on the calculations, the angle through which the tire rotates is equal to 4.26 radians and 244.0 degrees.

<h3>How to calculate the angle?</h3>

In Physics, the distance covered by an object in circular motion can be calculated by using this formula:

S = rθ

<u>Where:</u>

  • r is the radius of a circular path.
  • θ is the angle measured in radians.
  • S is the distance.

Substituting the given parameters into the formula, we have;

1.87 = 0.44 × θ

θ = 1.87/0.44

θ = 4.26 radians.

Next, we would convert this value in radians to degrees:

θ = 4.26 × 180/π

θ = 4.26 × 180/3.142

θ = 244.0 degrees.

Read more on radians here: brainly.com/question/19758686

#SPJ1

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Her resultant displacement is (45 Meters East.)

She originally walks 45 meters east, then she goes south 45 meters, then north 45 meters.  The south and north placements  just make her go back to where her previous placements were before them. Making her 45 meters east.

6 0
3 years ago
A 30 g particle is undergoing simple harmonic motion with an amplitude of 2.0 ✕ 10-3 m and a maximum acceleration of magnitude 8
KiRa [710]

Answer:

Explanation:

By using the Newton second law and the position equation for a simple harmonic motion we have

F=ma\\a_{max}=\omega^{2}A\\x=Acos(\omega t+ \phi)\\

where a is the acceleration, w is the angular frecuency and \phi is the phase constant. We can calculate w from the equation for the maximum acceleration

\omega=\sqrt{\frac{a_{max}}{A}}=\sqrt{\frac{8*10^{3}m/s^{2}}{2*10^{-3}}}=2000rad/s

(a).

F=ma=m\omega^{2}Acos(\omega t + \phi)\\F=(30*10^{-3}kg)(2*10^{-3}m)(2000\frac{rad}{s})^{2}cos(2000\frac{rad}{s} t - \frac{\pi}{2})=240N*cos(2000\frac{rad}{s} t - \frac{\pi}{2})

(b). T=\frac{2\pi}{\omega}=\frac{2\pi}{2000}=3.14*10^{-13} s

(c). v_{max}=A\omega=(2*10^{-3}m)(2000\frac{rad}{s})=4\frac{m}{s}

(d). The mecanical energy is the kinetic energy when the velocity is a maximum

E_{m}=E_{k}(v_{max})=\frac{mv_{max}^{2}}{2}=\frac{30*10^{-3}kg(4\frac{m}{s})^{2}}{2}=0.024J

3 0
3 years ago
Read 2 more answers
6) If nuclei A has a probability P of decaying in time t and nucleus B has a probability 2p of decaying in time t, which stateme
vredina [299]

Answer:

sorry dont know so so sorry

6 0
3 years ago
Question 1 Water: Start with the Water tab. Note that light areas represent places where the water is high (crests). Dark areas
Andre45 [30]

Answer:

Looks like a cosine function graph. The wave pattern is transversal waves . The faster the amplitude the higher the wave. The force of the drop hitting the water pushes the water down and out causing waves  If the water hits from a higher amplitude the waves raise bigger . When you increase the frequency of the water drops the waves move faster but no bigger.  When you increase the frequency of the water drops the wave ripples faster .

Explanation:

7 0
3 years ago
50. A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 335 kV. The secondar
Sedaia [141]

Answer:

Explanation:

a )  The transformer steps  up  the voltage from 12000 V  to 335000 V . Voltage in primary is 12000 V and in the secondary it is 335000 V in old transformer

If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils

the formula is

n₂ / n₁ = voltage in secondry / voltage in primary

n₂ / n₁ = 335000 / 12000

ratio of turns  in old transformer

= 27.9

ratio of turns  in new transformer

n₃ / n₁ = 750 / 12 ( n₃ is no of turns in the  secondary of new transformer )

= 62.5

T he ratio of turns in the new secondary compared with the old secondary

n₃ / n₂ = 62.5 / 27.9

= 2.24

b ) Current in secondary / current in primary

= turns in primary / turns in secondary

current output ratio of old

= Current in secondary / current in primary

= n₁ / n₂

= 12 / 335

= .03582

current output ratio of new

= Current in secondary / current in primary

= n₁ / n₃

= 12 / 750  

= .016

The ratio of new current output to old output (at 335 kV) for the same power

= .016 / .03582

= .4466

c ) power loss in new

=  (current in secondary )² x resistance of secondary

=( .016 x current in primary )² x R

= 2.56 X 10⁻⁴ X ( current in primary )² x R

power loss in old  

=  (current in secondary )² x resistance of secondary

=( .03582 x current in primary )² x R

= 12.83 X 10⁻⁴ X ( current in primary )² x R

ratio of new line power loss to old

= 2.56 / 12.83

= .199

7 0
3 years ago
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