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Yakvenalex [24]
3 years ago
11

A bend in a river shaped like a loop is called a(n) ________________

Physics
2 answers:
Llana [10]3 years ago
8 0
This loop-like bend in a river is called a meander
babymother [125]3 years ago
5 0
Meander

I think that would be it.
You might be interested in
How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?
Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

d =  \frac{ \:Δv}{e}

=  \frac{1.5v}{100v/m}

= 1.5 \times 10 {}^{ - 2} m

learn more about potential difference from here: brainly.com/question/28166044

#SPJ4

6 0
11 months ago
If yall friend me ill give lots of points like this in the future
Ne4ueva [31]

Answer:

ok......................

3 0
2 years ago
A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ
emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

5 0
3 years ago
A human being can be electrocuted if a current as small as 51.0 ma passes near the heart. an electrician working with sweaty han
boyakko [2]

The fatal current is 51 mA = 0.051 Ampere.

The resistance is 2,050Ω .

Voltage = (current) x (resistance)

            =  (0.051 Ampere) x (2,050 Ω)  =  104.6 volts .

==================

This is what the arithmetic says IF the information in the question
is correct.

I don't know how true this is, and I certainly don't plan to test it,
but I have read that a current as small as  15 mA  through the
heart can be fatal, not  51 mA .

If 15 mA can do it, and the sweaty electrician's resistance is
really 2,050 Ω, then the fatal voltage could be as little as  31 volts !

The voltage at the wall-outlets in your house is  120 volts in the USA !
THAT's why you don't want to stick paper clips or a screwdriver into
outlets, and why you want to cover unused outlets with plastic plugs
if there are babies crawling around.
6 0
3 years ago
A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/
kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 16.0 kg is the mass of the canoe

u_1 = -12.5 m/s is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

v_1 is the final velocity of the canoe

m_2 = 14.0 kg is the mass of the raft

u_2 = +16.0 m/s is the initial velocity of the raft

v_2 = -14.4 m/s is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s

So, the canoe is moving at 14.1 m/s to the right after the collision.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

5 0
3 years ago
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