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2 years ago

Which shows the weight of an atom?

Physics

2 answers:

harina [27]2 years ago

8 0

Answer:

Atomic mass

Explanation:

Dmitry_Shevchenko [17]2 years ago

7 0

Answer:

atomic mass

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Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

5 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

Now refer the schematic for visualization of situation:

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

Now the direction of the desired position with respect to south:

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

Now the distance from the current position to the desired position:

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

4 0

3 years ago

How do the chemical properties of the halogens compare to those of the noble gases?

serg [7]

Halogens are extremely reactive elements because they need one more electron to gain a full octet of valence electrons, whereas the noble gasesare extremely unstable because they already have their full octet.

8 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

3 years ago

Read 2 more answers

The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

5 0

3 years ago

Read 2 more answers