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viktelen [127]
3 years ago
14

A 35 kg child moves with uniform circular motion while riding a horse on a carousel. The horse is 3.2 m from the carousel's axis

of rotation and has a tangential speed of 2.6 m/s.
What is the child's centripetal acceleration?

Physics
2 answers:
musickatia [10]3 years ago
7 0
A=ω²r=0.8²*3.2=2.0 m/s²
v=ωr, then ω=v/r;
ω=2.6/3.2=0.8 rad/s
Lana71 [14]3 years ago
4 0

The answer is 74 N on edge.

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An electron enters a region of space containing a uniform 0.0000193-T magnetic field. Its speed is 121 m/s and it enters perpend
Mandarinka [93]

Answer

Given,

Magnetic field, B = 0.0000193 T

speed, v = 121 m/s

mass of electron, m = 9.11 x 10⁻³¹ Kg

charge of electron, q = 1.6 x 10⁻¹⁹ C

radius of the electron path, r = ?

r = \dfrac{mv}{qB}

r = \dfrac{9.31\times 10^{-31}\times 121}{1.6\times 10^{-19}\times 0.0000193}

r = 3.64 x 10⁻⁵ m

We know frequency is inverse of time period

d = v t

2 \pi r = v \times t

t = \dfrac{2 \pi r}{v}

t = \dfrac{2 \pi \times 3.64\times 10^{-5}}{121}

t = 1.889 x 10⁻⁶ s.

now, frequency

f = \dfrac{1}{t}

f = \dfrac{1}{1.889\times 10^{-6}}

f = 529380\ Hz

3 0
3 years ago
You have a horizontal cathode ray tube (CRT) for which the controls have been adjusted such that the electron beam should make a
Dvinal [7]

Answer:

Explanation:

Rotate it slowly to establish whether the spot moves. If it is broken, it will stay as it is. By rotating it the electrons' path to the centre of the screen will be tilted to another position by any external disturbing field.

8 0
3 years ago
ASAP need physics help please:)
boyakko [2]

Answer:

plantation of force of the earth acting on 15 kg of object on free fall the acceleration of free fall said about it now it is year 15 kg then it becomes 1.53 Newton

3 0
3 years ago
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OlgaM077 [116]

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4 0
3 years ago
A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the
Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

           due to the movement in opposite direction)

           Change in time period = Δt = 1.50 s

<u>Required</u>:

           Change in angular velocity = Δω = ?

           Average angular acceleration = αav = ?

<u>Solution</u>:

          <u>Angular velocity (Δω):</u>

               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0
3 years ago
Read 2 more answers
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