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klio [65]
3 years ago
12

A ball attached to a string is whirled around in a horizontal circle having a radius r. If the radius of the circle is changed t

o 4r and the same centripetal force is applied by the string, the new speed of the ball is which of the following?a. One-quarter the original speed
b. Twice the original speed
c. One-half the original speed
d. Four times the original speed
Physics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

Option D: Four times the original speed.

Explanation:

A centripetal force accelerates a body by changing the direction of the body's  velocity without changing the body's speed.

The speed(v) is therefore constant, thereby making the magnitudes of the of the acceleration and the force constant.

The formula used to calculate the Centripetal force is given below:

F = \frac{mv^2}{r}

where F represents the Centripetal force, m represents the mass of the moving body, v represents the speed or velocity at which the body is moving and r represents the radius.

Making the speed the subject of the formula: v^{2} = \frac{rF}{m} \\

Therefore, when the radius (r) is changed to 4r, i.e r = 4r

speed(v) becomes v^2 = \frac{4rF}{m}

After comparing, the difference between the speeds is Four times the original speed.

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Answer:

857.5 m

2.8583×10⁻⁶ seconds

Explanation:

Time taken by the sound of the thunder to reach the student = 2.5 s

Speed of sound in air is 343 m/s

Speed of light is 3×10⁸ m/s

Distance travelled by the sound = Time taken by the sound × Speed of sound in air

⇒Distance travelled by the sound = 2.5×343 = 857.5 m

⇒Distance travelled by the sound = 857.5 m

Time taken by light = Distance the light travelled / Speed of light

\text{Time taken by light}=\frac{857.5}{3\times 10^8}\\\Rightarrow  \text{Time taken by light}=2.8583\times 10^{-6}

Time taken by light = 2.8583×10⁻⁶ seconds

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Rank these significant figures numbers from the least to the most
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8 0
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A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o
lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

K_{1} + W_{F} = K_{2}

Where:

W_{F} - Work done by the external force, measured in joules.

K_{1}, K_{2} - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

W_{F} = K_{2} - K_{1}

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})

Where:

m - Mass of the object, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}

v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}

v_{1} \approx 25.060\,\frac{m}{s}

Final velocity

v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}

v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}

v_{2} \approx 33.121\,\frac{m}{s}

Finally, if m = 3.5\,kg, v_{1} \approx 25.060\,\frac{m}{s} and v_{2} \approx 33.121\,\frac{m}{s}, then the work done by the force is:

W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]

W_{F} = 820.745\,J

The work done by the force is 820.745 joules.

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Clouds may act to either reflect energy out to space or absorb energy, trapping it in the atmosphere.
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