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3 months ago

15

An object is traveling at a constant velocity of 15 m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11

s. What will its velocity be after that acceleration?

1 answer:

serg [3.5K]3 months ago

7 0

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

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In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c

Sav [3153]

Answer:

31.4 mm²

Explanation:

The ability of a telescope or eye to gather light can be expressed by the formula,

$GDP=\pi } \frac{d^{2} }{4}$

where d signifies the diameter of the pupil.

In bright daylight, the usual size of the pupil is 3 mm.

$GDP_{b} =\pi \frac{3^{2} }{4}$

Conversely, in darkness, the diameter typically enlarges to 7 mm.

$GDP_{b} =\pi \frac{7^{2} }{4}$

This indicates an increase in light-gathering capacity.

$Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}$

Thus, the amount of light the eye can capture is 31.4 mm².

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3 months ago

A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon

ValentinkaMS [3465]

When the boat submerges completely in the pond, the water level of the pond rises.

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2 months ago

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

Maru [3345]

Answer:

a) $X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b) D does not influence the long-term results.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$ This is a linear equation hence the integration factor, I

$I=e^{\int kdt}$

$I=e^{kt}$ Now using the characteristics of linear equations

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

b) At t= 0

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

Thus, the initial condition

does not affect the long-term outcome.

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

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3 months ago

An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the oc

Ostrovityanka [3204]

Answer:

F=126339.5N

Explanation:

To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:

Fw=W+Fi+F

where

Fw= force or weight exerted by the water column above the submarine.

To calculate Fw, we can use:

Fw=h. γ. A

h=height

γ= specific weight of seawater = 10074N / m ^ 3

A=Area

Fw=28x10074x0.7=197467N

w represents the hatch weight = 200N

Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:

Fi=PA=101325x0.7=70927.5N

Finally, the force needed to open the hatch is determined by the original equation:

Fw=W+Fi+F

F=Fw-W+Fi

F=197467N-200N-70927.5N

F=126339.5N

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4 months ago

A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

Sav [3153]

Answer:

The water level increases more when the cube is above the raft before it sinks.

Explanation:

The principle involved is based on Archimedes' theory, which states that immersing an object in water will raise the initial water level. This means that the height of the water in the container increases. The increase in water volume corresponds to the volume of the submerged object.

We can consider two scenarios: when the steel cube rests on the raft and when it settles at the pool's bottom.

When the cube rests at the pool’s bottom, the volume will indeed rise, and we can ascertain this using the cube's volume.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

When an object floats, it's because the densities of the object and water are in equilibrium.

$Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]$

The formula for density is:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyant force can be calculated with this equation:

$F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]$

Vs > Vc indicates that the combined volume of the raft and the cube exceeds that of the cube alone resting at the bottom of the pool. Hence, when the cube is positioned above the raft, the water level rises more before it becomes submerged.

7 0

4 months ago