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2 years ago

14

Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1m x 1m x 0.10m slab of foam, so

that the slab of foam floats in the water, leaving the bricks dry?

1 answer:

goblinko [34]2 years ago

4 0

Answer: Weight = 98.1N

Explanation:

Density of water = 1000 kg/m^3

Given that the Plastic foam is about 0.10 times as dense as water. That is,

Density of plastic foam = 0.1 × 1000 = 100kg/m^3

The volume V = 1 ×1×0.1 = 0.1 m^3

Density is the ratio of mass to volume

Density = mass/volume

Let us substitute for density and volume to get mass.

100 = M/0.1

Make M the subject of formula

M = 100 × 0.1 = 10 kg

Weight = mg

Where g = 9.81 m/s

Substitute the M and g into the formula

Weight = 10 × 9.81 = 98.1 N

Therefore, the weight of the brick is 98.1 N

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A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

$v=r\omega$

Kinetic energy is given by

$KE=0.5I\omega^{2}$

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

$0.5I\omega^{2}=mgh$ where m is mass, I is moment of inertia, $\omega$ is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, $\frac {2v}{L}$ for $\omega$ we obtain

$0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)$

$(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I}$ and making v the subject

$v^{2}=\frac {gl^{3}(m2-m1)}{4I}$

$v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}$

For the rod, moment of inertia $I=\frac {ML^{2}}{12}$ and for sphere $I=MR^{2}$ hence substituting 0.5L for R then $I=M(0.5L)^{2}$

For the sphere on the left hand side, moment of inertia I

$I=m1(0.5L)^{2}$ while for the sphere on right hand side, $I=m2(0.5L)^{2}$

The total moment of inertia is therefore given by adding

$I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}$

Substituting $\frac {L^{2}(M+3m1+3m2)}{12}$ for I in the equation $v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}$

Then we obtain

$v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}$

This is the expression of linear speed. Substituting values given we get

$v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s$

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2 years ago

Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o

mel-nik [20]

Answer:

a. 3/4λ

d. 1/4λ

Explanation:

When the wavelength of the sound waves is λ and the two waves are having same frequency the waves are said to be out of phase if their phase difference is in the multiples of $\frac{\lambda}{2}$ or 180°.

When the two waves are out of phase then their opposite maxima coincide at the same time resulting in the minimum amplitude of the resulting wave throughout.

As we observe from the schematic that the a wave has sinusoidal pattern of variation and we get a maxima after each $\frac{\lambda}{4}$ of the distance.

Here we have two speakers out of phase therefore on shifting one of the speakers by the odd multiples of $\frac{\lambda}{2}$ we have the maxima or the extreme amplitudes.

So, we must place the microphone at 3/4λ and 1/4λ to pickup the loudest sound.

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2 years ago

Two swimmers begin a race and Swimmer A completes each length of the pool in 30 seconds, while Swimmer B completes each length i

Anni [7]

Answer:1.33 times

Explanation:

Given

A takes 30 sec to complete each length

While B takes 40 sec to complete each length

Let L be the Length of Swimming pool Length

and $V_a$ and $V_b$ be the length of A & B

Thus after 5 minutes A traveled

$x_a=\frac{L}{30}\times 5\times 60=10L$

$x_b=\frac{L}{40}\times 5\times 60=7.5L$

so A is 2.5L units ahead of B

A has traveled 1.33 times than B in 5 mins

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You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm

maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

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A roller coaster travels 200 feet horizontally and then rises 135 feet at an angle of 30 degrees above the ground. What is the m

Delvig [45]

Say the initial point is (0,0)

The final point is

x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft

Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft

8 0

2 years ago