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jeka94
3 years ago
9

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi

th the clock to the distant planet CornTeen and measure the period of the clock to be 4s. By what factor is the gravitational acceleration constant g different on planet CornTeen compared to g on Earth
Physics
1 answer:
Elenna [48]3 years ago
5 0

Answer:

\frac{g_{2}}{g_{1}} = \frac{1}{4}

Explanation:

The period of the simple pendulum is:

T = 2\pi\cdot \sqrt{\frac{l}{g} }

Where:

l - Cord length, in m.

g - Gravity constant, in \frac{m}{s^{2}}.

Given that the same pendulum is test on each planet, the following relation is formed:

T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}

\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}

\frac{g_{2}}{g_{1}} = \frac{1}{4}

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With a circle.

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Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend
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Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

3 0
3 years ago
A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it
Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface \Phi it is possible to calculate the magnitude \frac{\Phi}{A}=B\ cos \theta.

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux \Phi is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is (m^{2}). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with \theta being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

                                                     \Phi=B\ A\ cos\theta

We are told the values of \Phi and B, then we can calculate the magnitude

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3 0
2 years ago
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject
kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

Radius_{mean} = 60\ mm

Diameter_{mean} = 120\ mm

Load = 200 N

We need to calculate the deflection

Using formula of deflection

\delta=\dfrac{8pD^3n}{Cd^4}

Put the value into the formula

\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}

\delta =34.56\ mm

Hence, The deflection of the spring is 34.56 mm.

4 0
3 years ago
The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a
vampirchik [111]

Answer:

F = 1.5 \times 10^{-16} N

this force is 1.68 \times 10^{13} times more than the gravitational force

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KE = 1 keV

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KE = 1.6 \times 10^{-16} J

now the speed of electron is given as

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F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})

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Now if this force is compared by the gravitational force on the electron then it is

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\frac{F}{F_g} = 1.68 \times 10^{13}

so this force is 1.68 \times 10^{13} times more than the gravitational force

4 0
3 years ago
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