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2 years ago

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

Physics

2 answers:

pantera1 [17]2 years ago

8 0

Answer:

this is under the wrong section

kotegsom [21]2 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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I’ll literlly give you points <br> and make you brainiest

sesenic [268]

Answer:

Explanation:

4 0

2 years ago

A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?

12345 [234]

The block is made of A) Tin, as its specific heat capacity is $0.225 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by $\Delta T$ , according to the following equation :

$Q=mC_s \Delta T$

where $C_s$ is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

$Q = 450 J$ is the amount of energy supplied to the block

$\Delta T = 1^{\circ}C$ is the change in temperature of the material

Solving the equation for $C_s$ , we can find the specific heat capacity of the unknown sample:

$C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)$

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

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7 0

2 years ago

Plz help me asap u will get a new friend

emmainna [20.7K]

2 is sedimentary and 3 is metamorphic

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2 years ago

The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

2 years ago

water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49

vesna_86 [32]

When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
$\theta = \arcsin ( \frac{n_2}{n_1} )$
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
$\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}$

6 0

2 years ago