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murzikaleks [220]

3 years ago

11

If the position of a particle on the x-axis at time t is âˆ’5t2, then the average velocity of the particle for 0 â‰¤ t â‰¤ 3 is

1 answer:

Drupady [299]3 years ago

3 0

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

$v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }$

let's look for the displacements

t = 0 x₀ = 0 m

t = 3 $x_{f}$ = 5 3 2

x_{f} = 45 m

we substitute

$v = \frac{45 -0}{3 - 0}$

v = 15 m / s

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Answer:

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2 years ago

A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse

bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, $f=4.2\times 10^{14}\ Hz$

Time, $t=3.2\times 10^{-11}\ s$

(a) Let $\lambda$ is the wavelength of this light. It can be calculated as :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}$

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or

$\lambda=714\ nm$

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

$n=f\times t$

$n=4.2\times 10^{14}\times 3.2\times 10^{-11}$

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Hence, this is the required solution.

8 0

2 years ago

Infer why the output force exerted by a rake must be less than input force?

adell [148]

<h3><u>Answer and Explanation</u>;</h3>

input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
<em><u>The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. </u></em>
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5 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

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Rudik [331]

Answer:

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2 years ago