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olganol [36]
3 years ago
11

If a 1.50 kg mass revolves at the end of a string 0.50 m long, and its tangential speed is 6.0 m/s, calculate the centripetal fo

rce.
Physics
2 answers:
Allisa [31]3 years ago
8 0

Answer:

108\ N

Explanation:

Mass of object, m=1.50\ kg.

Length of string, r=0.50\ m.

Tangential speed, v_t=6.0\ m/s.

Now, centripetal force F of a object moving in given radius r and mass m moving with velocity v.

Here , object moves along the ends of string. Therefore, radius is equal to length of string.

F=\dfrac{m \times v_t^2}{r}.

Putting values of m,v and r in above equation.

We get, F=\dfrac{1.50\times (6.0)^2}{0.50} \ N=108 \ N.

Hence, this is the required solution.

rosijanka [135]3 years ago
5 0
You should get about 110 for an answer
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Most likely, what causes the cylinder head temperature and engine oil temperature gauges to exceed their normal operating ranges is using fuel that has a lower-than-specified fuel rating. This can lead to detonation of the engine which is the tendency for the fuel to pre-ignite or auto-ignite in an engine's combustion chamber.The cylinder head and the engine oil are part of the automobile systems that helps in fuel combustion.

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Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from
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2 years ago
A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr
Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of 34.9^{\circ}

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A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

d sin \theta = m \lambda

where

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For laser 1,

d sin \theta = m_1 \lambda_1

For laser 2,

d sin \theta = m_2 \lambda_2

where

\lambda_1 = 420 nm\\\lambda_2 = 630 nm

Since the position of the maxima in the two cases overlaps, then the term d sin \theta on the left is the same for the two cases, therefore we can write:

m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}

Therefore:

m_1 = 3

m_2 = 2

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

d sin \theta = m_1 \lambda_1

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m

m_1 = 3 is the order of the maximum

\lambda_1 = 420 nm = 420\cdot 10^{-9} m is the wavelength of the laser light

Solving for \theta, we find the angle of the maximum:

sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572

So the angle is

\theta=sin^{-1}(0.572)=34.9^{\circ}

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

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