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3 years ago

Unweathered rock that underlies soil is subsoil. True False

2 answers:

kondor19780726 [428]3 years ago

8 0

This is false; rock that lies under soil is not known as subsoil, it is known as bedrock; a solid underlying loose deposits, such as soil.

Maru [420]3 years ago

5 0

False.

Unweathered rock that underlies soul is known as bedrock.

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Who has the best answer for this

ioda

Answer:

electricity because it has more percentage nd energy

Explanation:

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6 0

2 years ago

Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.50 kg mass to the

frez [133]

Given: Mass m = 0.50 Kg; Force = Weight = mg F = (0.50 Kg)(9.8 m/s²)

F = 4.9 N

Displacement x = 3.0 cm convert to meter   x = 0.03 m

Required: Spring constant k = "

Formula: F = kx

k = F/x

k = 4.9 N/0.03 m

k = 163.33 N/m

3 0

3 years ago

Describe where metals and nonmetals are found on the periodic table

Studentka2010 [4]

Non metals are all the way on the right side

Metalloids are the stair case

Metals are on the left side

4 0

3 years ago

Tungsten has a temperature coefficient of resistivity of 0.0045 (c°)-1. a tungsten wire is connected to a source of constant vol

Murljashka [212]

Answer:

$131.1^{\circ}C$

Explanation:

The power delivered in the wire is given by:

$P=\frac{V^2}{R}$

where V is the voltage of the battery and R is the resistance of the wire.

Since the voltage of the battery is constant, we can rewrite this equation as follows:

$V^2 = PR=const.$ (1)

At the beginning, the initial resistance is $R_0$ , and the power delivered is $P_0$ . Later, when the temperature increases, the power becomes $P_1 = \frac{2}{3}P_0$ , and the new resistance is $R_1$ . Using (1), we can write

$P_0 R_0 = \frac{2}{3}P_0 R_1\\R_1 = \frac{3}{2}\frac{P_0 R_0}{P_0}=\frac{3}{2}R_0$ (2)

So, the new resistance must be 3/2 of the initial resistance.

We know that the resistance increases linearly with the temperature, as

$R_1 = R_0 (1+\alpha \Delta T)$

where

$\alpha = 0.0045 ^{\circ}C^{-1}$ is the temperature coefficient

$\Delta T$ is the change in temperature

Using (2), we can rewrite this equation as

$\frac{3}{2}R_0 = R_0(1+ \alpha \Delta T)$

and we find:

$\frac{3}{2}=1+\alpha \Delta T\\\Delta T=\frac{\frac{3}{2}-1}{\alpha}=111.1 ^{\circ}$

So, the new temperature of the wire must be

$T_f = 21^{\circ}+111.1^{\circ}=132.1^{\circ}$

6 0

3 years ago

A stone whirled at the end of the a rope 30cm long, makes 10 complete resolution in 2 seconds Find: A. The angular velocity in r

lord [1]

Answers:

A: Angular velocity $\omega=31.40 \frac{r a d}{s}$

B: Linear velocity $v=9.42 \frac{m}{s}$

C: Linear Distance $d=47.1 \mathrm{m}$

Given:

Radius of the rope r=30cm=0.3m

Angular distance $\Delta \theta$ =10 revolutions

Time taken t=2seconds

To find:

A: Angular velocity in radians

B: Linear speed

C: Distance covered in 5 seconds

<u>Step by Step Explanations:</u>

Solution:

A: Angular velocity in radians;

According to the formula, Angular velocity can be calculated as

Angular Velocity = angular distance/ time

$\omega=\Delta \theta / \Delta t$

Where $\omega$ =Angular velocity

$\Delta \theta$ =Angular distance=10 revolutions

Changing revolutions to radians multiply with $2 \pi$ , so that we get

$=10 \times 2 \pi$

$=10 \times 2(3.14)$

=62.80 rad/rev

$\Delta t$ =Change in time

Substitute the known values in the above equation we get

$\omega$ =62.80 / 2

$\omega=31.40 \frac{r a d}{s}$

B. Linear speed of the rope;

As per the formula

Linear speed = angular speed × radius

$v=\omega \times r$

Where $\omega$ =Angular velocity

v=Linear speed of the rope

r=Radius of the rope

Substitute the known values in the above equation we get

$v=31.40 \times 0.30$

$v=9.42 \frac{m}{s}$

C. Dsitance covered in 5 seconds;

Linear distance = linear speed × time

$d=v \times t$

Where d= Linear distance of the rope

v=Linear speed of the rope

t=Time taken

Substitute the known values in the above equation we get

$d=9.42 \times 5$

$d=47.1 \mathrm{m}$

Result:

Thus A: Angular velocity of the rope $\omega=31.40 \frac{r a d}{s}$

B Linear speed of the rope $v=9.42 \frac{m}{s}$

C: Distance covered in 5 seconds $d=47.1 \mathrm{m}$

6 0

3 years ago