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Snowcat [4.5K]
3 years ago
10

How might the velocity (speed) of wind or water affect the deposition of sediments?

Physics
1 answer:
Maksim231197 [3]3 years ago
6 0
Well depending on the speed of both of those things is were the rock will be placed and it also determines how fast can an environment change
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Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of
maria [59]
P=m x v =60 x 7.0= 420
total momentum = 420 x 2,000,000,000=dnt have a calculator
thier effect would shake the earth and kill some organism because they jump up at the same time and they wil probably land the same time .
8 0
3 years ago
Read 2 more answers
Which are ways to improve the design of this experiment? Check all that apply.
Arte-miy333 [17]

Answer:

* Experiment with a higher range of materials

* Use a galvanometer.

* Vary in number of coils of the electromagnet

Explanation:

This is an experiment of electricity and magnetism, in general the best way to improve the results are:

* Experiment with a higher range of materials

   allowing to know the scope of the initial assumptions

* Use a galvanometer.

  The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.

* Vary in number of coils of the electromagnet

  Since it allows to have greater magnetic fields and therefore expand the range of measurements

3 0
3 years ago
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A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi
EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0
1 year ago
Please help!!!! Will mark brainliest.
julia-pushkina [17]

Answer:

Approximately 8.4 \times 10^{2}\; \rm N, assuming that g = 9.8\; \rm m \cdot s^{-2}.

Explanation:

Let m and a denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

  • Downward gravitational attraction from the earth: W = m \cdot g.
  • Upward tension force from the strand of web F(\text{tension}).

The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of F(\text{tension}) (which points upwards) should be greater than that of W (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

(\text{Net Force}) = F(\text{tension}) - W.

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

(\text{Net Force}) = m \cdot a.

Combine these two equations to get:

m \cdot a = (\text{Net Force}) = F(\text{tension}) - W.

Therefore:

\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}.

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately 8.4\times 10^{2}\; \rm N (downwards.)

4 0
3 years ago
14. Saeed is pulling a 6 kg heavy rock with an upward force of 40 N but does not succeed to lift it up. What is the magnitude of
NemiM [27]

Answer:

58.8 N

Explanation:

The normal force is calculated as equal to the perpendicular component of the gravitational force.

Thus; N = mg

We are given m = 6 kg

Thus;

N = 6 × 9.8

N = 58.8 N

Thus, magnitude of normal force on the rock = 58.8 N

6 0
2 years ago
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