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2 years ago

8

When a river overflows its banks, it deposits sediment over a broad, flat area of land on both sides of the river. This broad fl

at area is called a _______.
A.) channel
B.)delta
C.)valley
D.)floodplain

1 answer:

valentina_108 [34]2 years ago

6 0

The answer is letter D: a floodplain

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Consider heat transfer between two identical hot solid bodies and their environments. The first solid is dropped in a large cont

sergeinik [125]

<h2>For Second Solid Lumped System is Applicabe</h2>

Explanation:

Considering heat transfer between two identical hot solid bodies and their environments -

If the first solid is dropped in a large container filled with water, while the second one is allowed to cool naturally in the air than for second solid, the lumped system analysis more likely to be applicable

The reason is that a lumped system analysis is more likely to be applicable in the air than in water as the convection heat transfer coefficient so that the Biot number is less than or equal to 0.1 that is much smaller in air

Biot number = the ratio of conduction resistance within the body to convection resistance at the surface of the body

∴ For a lumped system analysis Biot number should be less than 0.1

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2 years ago

The wind blows a jay bird south with a force of 300 Newtons. The

adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The jay bird flies north, against the wind, with a force of 100 newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

$F=\sqrt{300^2+100^2} \\\\F=316.22\ N$

Hence, the net force on the jay bird is 316.22 N.

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2 years ago

A block is projected with speed v across a horizontal surface and slides to a stop due to friction. The same block is then proje

natka813 [3]

Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.

Explanation:

First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.

From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.

Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.

8 0

1 year ago

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre

alex41 [277]

Answer:

The simplified expression is $M = \frac{v^2 r}{G}$

Explanation:

From the question we are told that

$M = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }$

So simplifying we have

$M = \frac{m v^2}{r} * \frac{r^2 }{ mG }$

$M = \frac{v^2 r}{G}$

Thus the simplified formula is $M = \frac{v^2 r}{G}$

3 0

2 years ago

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kgÂ·m2. If the arms are pulled in so t

Serggg [28]

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular speed

Since the angular momentum must be conserved, we can write

$L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2$

where we have

$I_1 = 2.25 kg m^2$ is the initial moment of inertia

$\omega_1 = 5.00 rad/s$ is the initial angular speed

$I_2 = 2.25 kg m^2$ is the final moment of inertia

$\omega_2$ is the final angular speed

Solving for $\omega_2$ , we find

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s$

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

$K=\frac{1}{2}I\omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

$K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J$

- Final kinetic energy:

$K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J$

7 0

2 years ago