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3 years ago

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

1 answer:

7 0

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago

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A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th

ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a) Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0

3 years ago

A car, starting from rest, accelerates at

Mekhanik [1.2K]

Using V= vo +at with Vo = 0 and a= 4m/sec2.
V= 0+ 4x8= 32m/s

3 0

2 years ago

Read 2 more answers

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position $x_{i}=5.00\ cm$

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx$

Put the value into the formula

$\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}$

$v^2=\dfrac{2\times12.3285}{1.00}$

$v^2=24.657$

$v=4.96\ m/s$

Hence, The speed of the block is 4.96 m/s.

6 0

3 years ago

The picture below shows the setup for an experiment involving a 1000 mL beaker, sitting on a burner, filled with 500 mL of water

Vlad1618 [11]

Answer:

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Explanation:

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4 0

3 years ago

A positively charged sphere with a charge of 8Q is separated from a negatively charged sphere -2Q by a distance r. There is an a

djverab [1.8K]

Answer:

Explanation:

For the first case , the expression for electrostatic force can be given by the following .

F = K x 8Q x 2Q / r² where k is a constant .

F = K 16 Q² / r²

When they touch , some charge is neutralized . Net charge remaining

= 8Q - 2 Q = 6 Q

Charge on each sphere = 6Q/2 = 3 Q .

Force between them

F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²

F₁ / F = 9 / 16

F₁ = 9 F / 16 .

4 0

2 years ago