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3 years ago

8

Which is an example of natural erosion? ice forming in cracks of rocks acid rain falling on sidewalks waves washing over rocks o

n the beach water washing away soil in an area with off-road vehicles

2 answers:

ozzi3 years ago

7 0

Answer: waves washing over rocks on the beach

Explanation:

A natural erosion is a phenomena of removal of the top layer of the soil or any other surface material by the action of the natural physical agents like water, wind and others. Waves washing over rocks on the beach is the correct example of natural erosion because waves from any water body are naturally generated by the effect of wind and gravity these can wipe the surface materials present over the rocks on the beach.

oee [108]3 years ago

5 0

Ice forming in cracks of rocks

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In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

$s=ut+\dfrac{1}{2}at_{A}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

$t_{A}=\sqrt{\dfrac{30\times2}{1.6}}$

$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

4 0

4 years ago

When does acceleration due to gravity equal 9.8 m/s downward?

Elza [17]

Answer:

b

Explanation:

i took the quiz i think its right

7 0

3 years ago

Calculate Speed The 2-kg metal ball moving at a speed of 3 m/s strikes a 1-kg wooden ball that is at rest. After the collision,

enot [183]

Answer:

53466kg.

Explanatiokn:

5 0

2 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago

A pizza delivery driver must make three stops on her route. She will first leave the restaurant and travel 4 km due north to the

topjm [15]

<h2>5.3 km</h2>

Explanation:

This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.

Let us denote each of the individual displacements by a vector. Consider the unit vectors $\vec{i}\textrm{ and }\vec{j}$ as the unit vectors in the direction of East and North respectively.

By simple calculations, we can derive the unit vectors $\vec{j},\frac{-\vec{i}-\vec{j}}{2}\textrm{ and }\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2}$ in the directions North, $45^{o}$ South of West and $60^{o}$ North of West respectively.

So Total displacement vector = Sum of individual displacement vectors.

Displacement vector = $4(\vec{j})+6(\frac{-\vec{i}-\vec{j}}{2})+5(\frac{-\frac{1}{2}\vec{i}+\frac{\sqrt{3}}{2}\vec{j}}{2})=-4.25\vec{i}+3.165\vec{j}$

Magnitude of Displacement = $|-4.25\vec{i}+3.165\vec{j}|=5.3km$

∴ Total displacement = $5.3km$

4 0

3 years ago