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3 years ago

What conditions are typcial of the soil the deeper we dig into it?

1 answer:

8 0

When we first dig into soil, it's nice and soft.( Depending on what soil it is; usually topsoil)

But when we go deeper into it, there are rocks and the soil is harder. There is bedrock at the bottom, subsoil in the middle, and topsoil at the top, which is where we walk.

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Help me please i dont understand

alexandr402 [8]

Option A is the correct answer.

5 0

3 years ago

Three charges are arranged as shown in the picture above. Find the magnitude and direction of the electrostatic force on the 6 n

bagirrra123 [75]

Answer:

Bot Nm

Explanation:

jdjdmxjd mdjdcj jdsdj jedidj jddj

6 0

3 years ago

A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie

Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0

3 years ago

3. A car has a mass of 2.50 x 10^3 kg. If the force acting on the car is 7.65 x 10^3 N to the

const2013 [10]

Answer:

3.06m/s² to the east

Explanation:

Given parameters:

Mass of car = 2.5 x 10³kg

Force acting on the car = 7.65 x 10³N

Unknown:

Acceleration of the car = ?

Solution:

From Newton's second law of motion:

Force = mass x acceleration

Acceleration = $\frac{Force }{mass}$ = $\frac{7.65 x 10^{3} }{2.5 x 10^{3} }$ = 3.06m/s² to the east

6 0

3 years ago

What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to

Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

$F_G=\frac{Gm_1m_2}{r^2}$ where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

$F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2}$ = $\frac{2}{4} \frac{Gm_1m_2}{r^2}$

Therefore the gravitational force is halved. That is it will always decrease

4 0

3 years ago