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3 years ago

Alessandra jumps on a trampoline. It takes her 1 second to

Physics

1 answer:

svp [43]3 years ago

5 0

Answer:

The speed of Alessandra's motion is 2 ft/s up.

Explanation:

Given that,

Time = 1 sec

Height = 2 feet

We need to calculate the speed of Alessandra's motion

Using formula of speed

$v=\dfrac{d}{t}$

Where, v = speed

d = height

t = time

Put the value into the formula

$v=\dfrac{2}{1}$

$v=2\ ft/s$

The direction will be up ward.

Hence, The speed of Alessandra's motion is 2 ft/s up.

(C) is correct option.

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Which has a total mass of 1612 kg. If she accelerates from rest to a speed of 12.87 m/s in 3.47 s, what is the minimum power req

blondinia [14]

Answer:

76969.29 W

Explanation:

Applying,

P = F×v............. Equation 1

Where P = Power, F = force, v = velocity

But,

F = ma.......... Equation 2

Where m = mass, a = acceleration

Also,

a = (v-u)/t......... Equation 3

Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s

Substitute these values into equation 3

a = (12.87-0)/3.47

a = 3.71 m/s²

Also Given: m = 1612 kg

Substitute into equation 2

F = 1612(3.71)

F = 5980.52 N.

Finally,

Substitute into equation 1

P = 5980.52×12.87

P = 76969.29 W

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2 years ago

What objects can be seen from earth because they producde there own light? 

Lena [83]

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3 years ago

If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change

Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

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2 years ago

When light passes through a convex lens, it does this.....

Tpy6a [65]

Explanation:

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7. convex

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Read 2 more answers

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

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3 years ago