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3 years ago

Why is the inner core a dense ball of solid metal?

Physics

1 answer:

Zielflug [23.3K]3 years ago

4 0

Answer:

so dont mind this it just said the answer have to be 20 words long (the answer should be B)

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Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

6 0

3 years ago

Which of the following describes a risk associated with the use of hydropower as an energy source?

castortr0y [4]

Its A: the use of hydropower often changes the natural flow of water through an ecosystem

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5 0

3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

AnnyKZ [126]

Answer:

Time, $t=1.07\times 10^{-23}\ s$

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel $3.2\times 10^{-15}\ m$ before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

$t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s$

So, the time interval is $1.07\times 10^{-23}\ s$

5 0

3 years ago

Atoms of the same element that differ only in the number of neutrons are known as.

True [87]

Answer:

They are known as isotopes

7 0

3 years ago