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3 years ago

I need major help ! Write a hypothesis for a way to make a wave disappear.

Physics

1 answer:

stellarik [79]3 years ago

3 0

Answer:

A new theoretical design using nanowires provides a way to hide ... But to make things truly disappear before our eyes, a way will need to be ...

Shh! Computer simulations predict that certain materials could allow sound waves to pass by undisturbed, effectively making them acoustically ...

Explanation:

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Describe different types of motion with examples. (4)

Illusion [34]

Answer:

there r two types of motion, uniform and non-uniform

uniform means equal distance travelled at equal intervals of time

and non-uniform is exactly the opposite.

quantities which can be represented by magnitude along r called scalar quantities such as speed.

quantities which need magnitude along with direction r called vector quantities such as velocity.

velocity=10m/s

acceleration = u-v/s i.e initial final velocity - initial velocity upon time

acceleration= 0.2m/s sq

time= 30s

10 = displacement/time

10 = x/30

10 = 300

Answer is 300 meters - distance/displacement.

8 0

3 years ago

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A block of plastic in the shape of a rectangular solid that has height 8.00 cm and area A for its top and bottom surfaces is flo

kifflom [539]

Answer:

0.35 kg

Explanation:

8 cm = 0.08 m

For the block to stay balance, the buoyancy force must be the same as gravity that pulls it down.

Let mass of the block be M, then the gravity would be Mg

Let water density be $\rho_w = 1000 kg/m^3$ , the buoyancy force would be the weight of water that is displaced by the submerged block.

For example, when there is no coin, block is $h_0 = 0.0312m$ submerged. The weight of water displaced must be

$W_0 = Ah_0\rho_wg = 0.0312A1000g = 31.2Ag$

Which is also the weight of block, of Mg

Therefore M = 31.2A. (1)

As coins are stacked on top of block, h increase, so as weight of water displaced and total weight of block and coins. Now let m be the total weight of coins. The gravity of block and weight must be (M+m)g. And the weight of water displaced is:

$W = Ah\rho g = (M + m)g$

$h = \frac{M}{A\rho} + \frac{m}{A\rho}$

Since the linear plot of h vs m has a slope of 0.089 m/kg, we can interpret it as

$\frac{1}{A\rho} = 0.089$

$A = \frac{1}{0.089\rho} = \frac{1}{89} = 0.011 m^2$

So from the eq. (1) we can solve for M = 31.2A = 0.35 kg

8 0

3 years ago

Q1: Using the known values of the mass of the earth and the radius of the earth, calculate the

kupik [55]

Answer:

m g = G * m * M / R^2 force of attraction

g = G * M / R^2

g = 6.67E-11 * 5.96E24 / (6.37E6)^2

g = 6.67 * 5.96 / 6.37^2 * 10^1

g = .98 * 10 = 9.8 m/s^2

Q2.

d g = G M * (-2 R^-3 dR ) = -2 G M / R^3 (very small where dR = 1)

3 0

3 years ago

The Earth's magnetic field is approximately 0.5 Gauss. What is the magnetic energy density of this field

ser-zykov [4K]

Answer:

The magnetic energy density is $E_B = 9.95 *10^{-4} \ J/m^3$

Explanation:

From the question we are told that

The earths magnetic field is $B = 0.5 \ Gauss$

Now

$1 \ Gauss \to 1 *10^{-4} \ T$

=> $0.5 \ Gauss \to x T$

$x = \frac{0.5 * 1*10^{-4}}{1}$

$x = B= 5.0 *10^{-5} \ T$

Generally the magnetic energy density is mathematically represented as

$E_B = \frac{B^2}{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with a constant value

$\mu_o = 4\pi * 10^{-7} N/A^2$

substituting values for equation above

$E_B = \frac{(5*10^{-5})^2}{ 2 * 4\pi * 10^{-7} }$

$E_B = 9.95 *10^{-4} \ J/m^3$

5 0

3 years ago

Can someone help me with science:

Oliga [24]

Answer:

trur

Explanation:

gvidttcyhjddgdhjfjdjsjs

6 0

3 years ago

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