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4 years ago

5

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o

ffers a damping force numerically equal to Ï2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 5 ft/s. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?

1 answer:

Naddika [18.5K]4 years ago

6 0

a for s;lll the plkauito s;owernghadf vb vhebv3h

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Fine grains of beach sand are assumed to be spheres of radius 64.8 µm. These grains are made of silicon dioxide which has a dens

Nutka1998 [239]

Answer:

1) Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) 0.08901 kg of sand would have surface area equal the surface area of the cube.

Explanation:

1) Radius of the grain,r = 64.8 µm = $6.48\times 10^{-5}m$

Volume of the sphere = $\frac{4}{3}\pi r^3$

Volume of the grain of a sand:

$V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3$

$V=1.1397\times 10^{-12} m^3$

Density of a grain of sand = $d=2600 kg/m^3$

Mass of a grain of a sand = M

$d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}$

$M=2.9632\times 10^{-9} kg$

Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) Surface are of sphere: $4\pi r^2$

Surface area of a grain:

$A=4\times 3.14\times (6.48\times 10^{-5}m)^2$

$A=5.2766\times 10^{-8} m^2$

Length of the cube = a = 0.514 m

Total surface area of cube ,A'= $6a^2$

$A'=6\times (0.514 m)^2=1.5851 m^2$

let the number grains with area equal to total surface area of cube be x.

$A'=A\times x$

$x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7$

Volume of x number of grains :V'

$V'=V\times x$

$V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7$

$V'=3.4236\times 10^{-5} m^3$

Mass of $3.4236\times 10^{-5} m^3$ of sandL:

= $3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg$

0.08901 kg of sand would have surface area equal the surface area of the cube.

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3 years ago

Which describes an image that can be produced by a concave lens?

mixer [17]

A concave is when the bump is pressed downwards; So if the image has a downward hole, then it's concave. If it's pressed upwards like a hill or mountain, then it's convex.

5 0

4 years ago

What is significant figures

Maurinko [17]

Explanation:

Significant figures of a number written in positional notation are digits that carry meaningful contributions to the measurement resolution of the number.

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4 years ago

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Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

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4 years ago

I will mark as the brainliest answer<br><br>plz 8,9,10

blagie [28]

Answer:

8. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 unit.

10. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units.

Explanation:

8. i) acceleration = velocity / time

ii) In this figure velocity = time

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. i) acceleration = velocity / time

ii) In this figure 4 = m + 5, therefore m = -1

therefore velocity = (-0.5 $\times$ time) + 5

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 units.

10.) velocity is constant at 2

therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units

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4 years ago