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4 years ago

Which of the following is an example of biological weathering

1 answer:

3 0

An example of biological weathering is the widening of gaps in rocks caused by tree roots since pressure or stress is exerted on the rocks due to the growing plant roots. The pressure is exerted by a biological process (i.e., growing roots), even though the process is physical.

Plants and animals and its action speeds up mechanical and or chemical weathering, which in turn causes biological weathering.

<span>With the help of the chemicals that are produced by the tiniest bacteria, algae and liches, the rocks which they live in are also broken down by them in order to get the nutrients they need.</span>

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If the v – t graph of a particle is parallel to the t – axis the body has (a) uniform velocity (b) uniform acceleration (c) zero

maxonik [38]

Answer:

zero velocity

Explanation:

if the v-t graph is parallel to the time axis its mean body has covered no path in other words the body is at rest so the velocity of the body should be zero

hope my answer will helps u plz mark me brainlist

5 0

3 years ago

Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

ladessa [460]

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

$m_1v_1 + m_2v_2 = 0$

now plug in all data into above equation

$5(v) + 3(3)$

$5v = -9$

$v = -1.8 m/s$

so correct answer is

A) - 1.8 m/s

3 0

3 years ago

In which part of the scientific method do you sum up your results

elena-14-01-66 [18.8K]

Data Analysis and Conclusion

8 0

3 years ago

A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne

Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge $q_{1}=3.90\times10^{-12}\ C$

Negative charge $q_{2}=-2.60\times10^{-12}\ C$

We need to calculate the total charged

Using formula of charge

$Q_{enc}=2q_{1}+3q_{2}$

Put the value into the formula

$Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})$

$Q_{enc}=0$

We need to calculate the electric flux

Using formula of electric flux

$\phi=\dfrac{Q_{enc}}{\epsilon_{0}}$

Put the value into the formula

$\phi=\dfrac{0}{8.85\times10^{-12}}$

Hence, The electric flux is zero because charge is zero.

7 0

3 years ago

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago