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Nitella [24]
3 years ago
10

Josh has a toy car of mass 3 kg tied to a string of length 2 m. He ties the string to a pole and has the toy car drive in a circ

le around the pole at a speed of 3 m/s.
a. What is the centripetal acceleration of the car?

b. If the tension in the string exceeds 50 N, the string will break. How fast can he make the car go without breaking the string?
Physics
2 answers:
gulaghasi [49]3 years ago
5 0

Part a)

Centripetal acceleration is defined as

a_c = \frac{v^2}{R}

now here we know that

v = 3m/s

R = 2 m

m = 3 kg

now from above formula we have

a_c = \frac{3^2}{2}

a_c = 4.5 m/s^2

Part b)

Maximum possible tension in the string is given as

T = 50 N

now by force equation we have

F = ma

50 = 3 a

a = \frac{50}{3} m/s^2

now again by above formula

\frac{v^2}{R} = \frac{50}{3}

v = \sqrt{\frac{50 \times 2}{3}}

v = 5.77 m/s

aliina [53]3 years ago
5 0

a) a=v^2/R=4.5 m/s/s

b) F=ma=mv^2/R, so

v=\sqrt{FR/m}=5.77 m/s

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Answer: 115.52\ N

Explanation:

Given

Length of plank is 1.6 m

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\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N

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8 0
2 years ago
When air resistance equals the weight of an object, the object has reached
MissTica

Answer:

When air resistance equals the weight of an object, the object has reached free fall.

Explanation:

  • When an object has only force acting on it as gravity then, it experiences free fall.
  • During free fall all the forces except gravity is balanced by one another.
  • In the question, object's weight is balanced by air resistance so it is in the state of free fall.
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6 0
3 years ago
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
Elan Coil [88]

Answer:

Time period, T=3.05\times 10^{-5}\ s

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

T=\dfrac{1}{f}

T is the time period of the crystal's motion.

Time period is given by :

T=\dfrac{1}{32768}

T=3.05\times 10^{-5}\ s

So, the time period of the crystal's motion is 3.05\times 10^{-5}\ s. Hence, this is the required solution.

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