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2 years ago

5

A bowling pin is thrown vertically upward such that it rotates as it moves through the air, as shown in the figure. Initially, t

he center of mass of the bowling pin is moving upward with a speed vi of
10 The maximum height of the center of mass of the bowling pin is most nearly

1 answer:

zheka24 [161]2 years ago

3 0

Answer:

Explanation:

In projectile, the formula for calculating the maximum height reached by the pin is expressed as;

H = u²/2g

u is the speed/velocity = 10m/s

g is the acceleration due to gravity = 9.81m/s²

Substitute

H = 10²/2(9.81)

H = 100/19.62

H = 5.097m

Hence the maximum height of the center of mass of the bowling pin is most nearly 5.0m

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The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque

stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

$\lambda _1=589 nm$

$\lambda _2=589.6 nm$

Energy is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

x=velocity of Light

$\lambda$ =wavelength of emission

$E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}$

$E_1=3.374\times 10^{-19} J$

$E_1 in kJ/mol$

$E_1=203.2 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}$

$\nu _1=5.09\times 10^{14} Hz$

Energy corresponding to $\lambda _2$

$E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}$

$E_2=3.371\times 10^{-19} J$

$E_2=203.02 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}$

$\nu _1=5.088\times 10^{14} Hz$

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2 years ago

Radio waves travel 300,000,000 m/s. The frequency is 101,700,000. ehats the wavelength

Nesterboy [21]

Answer:

we have formula of frequency :

frequency(f)= speed of sound(c)/wavelength(λ)

for wavelength we swipe it with frequency as follows

λ=c/f

λ=300,000,000/101,700,000

λ=2.949

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2 years ago

sticking your fingers into a wall socket will not bring you into direct contact with an electrical current. true or false

SCORPION-xisa [38]

The correct answer is false cause how can u fit your finger in a wall something it's to small

3 0

2 years ago

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A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary

Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

<em>mass of the bullet, m = 23 g = 0.023 g</em>
<em>speed of the bullet, u = 230 m/s</em>
<em>mass of the wood, m = 2 kg</em>
<em>final speed of the bullet, v = 170 m/s</em>
<em>coefficient of friction, μ = 0.15</em>

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

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2 years ago

If the Earth got closer to the Sun, how would the gravity between the Sun and Earth change?

Neko [114]

Answer:

I think its 2. It would decrease

Explanation:

Hope this helps

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2 years ago

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