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2 years ago

5

A fighter plane is descending at 30 m/s. The pilot ejects, and the ejector seat accelerates him upwards at 120 m/s2 for 2 second

s. What is his velocity at the end of the 2 seconds?

1 answer:

katrin [286]2 years ago

4 0

Answer:

210m/s upwards

Explanation:

u= -30m /s

a= 120m/s^2

t=2s

v=u+at

=(-30)+(120x2)

=240-30

=210m/s

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Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time

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The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let $\overset \rightarrow {v_y}$ represent the velocity of the boat across the river in the direction, $\overset \rightarrow {d_y}$ , and let, $\overset \rightarrow {v_x}$ , represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = $\overset \rightarrow {v_y}$

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is $\overset \rightarrow {v_y}$ , such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

$\overset \rightarrow {v_y}$ = 3 m/s

$\overset \rightarrow {v_x}$ = 4 m/s

The width of the river, $w_y$ = 6 meters, we have;

The resultant velocity = $\sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s$

The direction, θ, is given as follows;

$\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}$

The length of the path of the boat, <em>l</em>, is given as follows;

$l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10$

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = $\dfrac{l}{v}$ , therefore;

$t = \dfrac{10}{5} = 2$

The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

$t = \dfrac{w_y}{v_y}$

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$t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s$

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Learn more vectors here:

brainly.com/question/15907242

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3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

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The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

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$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

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$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

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Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

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Answer:

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