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2 years ago

I need help with question two

Physics

1 answer:

zaharov [31]2 years ago

6 0

I can't see numbers here, so here are all the answers:
1) the frequency is c/λ = 3e8/556e-9 = 5.39e14Hz
2) light travels at <span>299,792,458 m/s. So in nanoseconds it's 0.299792458m. This is about 1/3 of a meter which is about one foot.
3) length is L = ct = (</span>299,792,458 m/s)(6e-15) = 1.799e-6m or 1.799μm

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What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?

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Hi there!

Recall Newton's Second Law:

$\large\boxed{\Sigma F = ma}$

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We must begin by solving for the acceleration using the following:

a = Δv/t

In this instance:

Δv = 3 m/s

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a = 3/2.5 = 1.2 m/s²

Now, plug this value along with the mass into the equation for net force:

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Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

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We need to calculate the time

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$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

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We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

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$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

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Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

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