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2 years ago

14

In the diagram, above, marker F is pointing to a __________, which are formed when meanders wear away at a narrow point and a po

ol of water forms.

2 answers:

castortr0y [4]2 years ago

4 0

The answer is <span>oxbow lake</span>

zubka84 [21]2 years ago

3 0

B. oxbow lake, gl on the program to you smart environmental scientists.

You might be interested in

Please help me with this i dont know it

Inessa05 [86]

Answer:

no it would not. that is an open circuit and it would need to be closed at the switch for current to flow.

5 0

2 years ago

Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 430 and 610 turns, respectively. A

KIM [24]

Answer

given,

Two solenoids A and B

Number of turn

Na = 430 turns Nb = 610 turns

Current = 2.80 A

Average flux through A = 300 μWb

Average of flux through B = 90.0 μ Wb

a) $L = \dfrac{N \phi}{I}$

$L = \dfrac{610\times 90 \times 10^{-6}}{2.80}$

$L =19.6 mH$

b) inductance of A

$L = \dfrac{N_A \phi_A}{I_A}$

$L = \dfrac{430\times 300 \times 10^{-6}}{2.80}$

$L =46 mH$

c) magnitude of the emf

$\epsilon_B = -L_B\dfrac{dI}{dT}$

$\epsilon_B = -(19.6\times 10^{-3})(0.5)$

$\epsilon_B = -9.8\times 10^{-3}\ V$

$\epsilon_B = -9.8\ mV$

7 0

2 years ago

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr

kolbaska11 [484]

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.

4 0

2 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

$P=\frac {W}{t}=\frac {FS}{t}$ where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328 W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

$F=\frac {P}{V}$

$F=\frac {125328}{23}=5449.043478 N$

F=5449 N

7 0

2 years ago

The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia

MAVERICK [17]

Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

= 1 J - 0.313 J

= 0.687 J

So,

Energy Lost for group A's car = 0.687 J

Now,

For Group B -

Energy Lost = GPE - KE

= 1 J - 0.45 J

= 0.55 J

So,

Energy Lost for group B's car = 0.55 J

8 0

2 years ago