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3 years ago

A pendulum that moves through its equilibrium position once every 1.000 s is sometimes called a "seconds pendulum."

1 answer:

3 0

<span>I would solve this problem by starting from the equation for a pendulum's period:

T ≈ 2π√( L / g )

Rearranging for g, and substituting the length in Cambridge:

g ≈ ( 4π² * L ) / T²
. . = ( 4π * 0.9942 m ) / ( 1.000 s + 1.000 s )²
. . = 9.812 m/s²

And for Tokyo:

g ≈ ( 4π² * L ) / T²
. . = ( 4π * 0.9927 m ) / ( 1.000 s + 1.000 s )²
. . = 9.798 m/s²</span>

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A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

4 0

4 years ago

Magnetic orientation in rocks that is opposite to the current orientation of Earth's magnetic field is called ____ magnetic pola

xxMikexx [17]

Answer:

Normal Magnetic Polarity

Explanation:

Normal Magnetic polarity.

The polarity can be either normal or reversed Natural polarity is where the north magnetic points (roughly) towards the north pole. ... The reversed polarity is in the opposite direction and the northern end of the magnetic field is near the current southern pole.

6 0

3 years ago

Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i

Sidana [21]

The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons

5 0

4 years ago

Teenagers

Sophie [7]

A would be the best

5 0

3 years ago

suppose a ball had a potential energy of 5 j when you dropped it. What would be its kinetic energy just as it hit the ground. (i

Andru [333]

Since we are ignoring air resistance which is a non-conservative force, the potential energy will be completely converted into kinetic energy, resulting in a final kinetic energy of $5J$ .

4 0

3 years ago