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Bond [772]
3 years ago
15

The distance, in feet, a moving object has traveled after t seconds is given by 2t/(4 + t). find the acceleration of the object

after 5 seconds. (round your answer to three decimal places.)
Physics
1 answer:
lianna [129]3 years ago
3 0
Hi! Let me help you!
a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0 
a = [(2t)/(4+t)]/t <---- working equation
a = {[2(5)]/9}/5 <---- cancel 5
a = 2/9 ft/s^2 <---- Answer
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The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou
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Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

 g = \dfrac{GM}{r^2}

 g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}

        g = 8.61 m/s²

3 0
3 years ago
you are boiling water to make pasta. How could you descibribe the pot, water, and pasta using physical properties?
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Answer:

Yes

Explanation:

As a personification

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A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top
elixir [45]

Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

8 0
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Describe a sea-level transgression
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LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

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3 years ago
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