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a_sh-v [17]
3 years ago
7

How is property belonging to individuals protected under the Fifth Amendment?

Physics
2 answers:
pantera1 [17]3 years ago
8 0

Answer:

this is under the wrong section

kotegsom [21]3 years ago
7 0

Answer:

By preventing the government from taking property without fair payment.

Explanation:

Answer 1

Please mark me as brilliant and thank you

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Drag the tiles to the correct boxes to complete the pairs. Match the sentences with the steps of the scientific method
Assoli18 [71]

Solution:

Make an Observation - An indoor plant in a dark room withers faster than the same plant in a room with ample sunlight.

Ask a question- Why do certain indoor plants die faster based on where they are placed in the house?

State a hypothesis- Sunlight is probably essential for plants to grow and live.

Run an experiment- Get two potted plants. Cover one with black paper. Place both plants outside in sunlight. See what happens to each plant after one week.

Analyze the results-The plant in the pot with black paper withered. The other plant was healthy.

Communicate the results to others - Plants need sunlight to make food so they can live.

4 0
3 years ago
What is the wavelength of a radio wave with a frequency of 2.00 x 10^4 Hz? (hint: what is the speed of all electromagnetic radia
patriot [66]

Answer: Option B

The wavelength for a radio wave with a frequency of  2 × 10⁴ Hz is

1.5 × 10⁴ m.

Explanation:

Wavelength is the measure of distance between two successive crests or troughs in a standing wave. Also wavelength can be measured as the ratio of velocity of light to frequency. It is like this because wavelength is inversely proportional to the frequency.

Wavelength =\frac{c}{frequency}

As c = 3 × 10⁸ m/s and the frequency is 2 × 10⁴ Hz, then the wavelength will be

Wavelength =\frac{3*10^{8} }{2*10^{4} } =1.5*10^{4} m

So, the wavelength for a radio wave with a frequency of  2 × 10⁴ Hz is

1.5 × 10⁴ m.

5 0
3 years ago
The children asked about the electric train and how to use it.
juin [17]
U use the train to play hope i helped please thank me 
3 0
3 years ago
The equations of kinematics describe the motion of an
Svetradugi [14.3K]

Answer:

The equations of kinematics is applied for the motion with constant acceleration (including zero), but the condition is that the acceleration should be in the direction of the motion (positive or negative).

In circular motion, the acceleration is radial (centripetal), which means that the acceleration is always perpendicular to the motion of the object, therefore the equations of kinematics cannot be applied.  

3 0
3 years ago
A mass m1 hangs from a spring k and is in static equilibrium. A second mass m2 drops through a height h and sticks to m1 without
jekas [21]

Answer:

\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}

Explanation:

From the information given:

The equation of the motion can be represented as:

(m_1 +m_2) \hat u + ku = m_2 g--- (1)

where:

m_1 = mass of the body 1

m_2 = mass of the body 2

\hat u = acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency \omega _n = \sqrt{\dfrac{k}{m_1+m_2}}

And the equation for the general solution can be represented  as:

u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)

To determine the initial velocity, we have:

\hat u_2^2 = 2gh

\hat u_2 = \sqrt{2gh}

where h = height

Suppose we differentiate equation (2) with respect to time t; we have the following illustration:

\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0

now if t = 0

Then

\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0

= \omega _n B

Using the  law of conservation of momentum on the impact;

m_2 \hat  u_2=(m_1+m_2) \hat u (0)

By replacing the value of \hat u_2 with \sqrt{2gh

Then the above equation becomes:

m_2 \times \sqrt{2gh}=(m_1+m_2) \ u(0)

Making u(0) the subject of the formula, we have:

u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}

Similarly, the value of the variable can be determined as follows;

Using boundary conditions

0 = A cos 0 + B sin 0 + \dfrac{m_2g}{k}

0 = A (1)+0+ \dfrac{m_2g}{k}

A =- \dfrac{m_2g}{k}

Also, if  \hat u (0) = \omega_nB

Then :

\dfrac{m_2}{m_1+m_2}\sqrt{2gh} = \omega_n B

making B the subject; we have:

B = \dfrac{m_2}{m_1 + m_2}\dfrac{\sqrt{2gh}}{\omega_n}

Finally, replacing the value of A and B back to the general solution at equation (2); we have the equation of the subsequent motion u(t) which is:

\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}

8 0
3 years ago
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