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Nostrana [21]
4 years ago
5

Why is it difficult to see colors outside at night?

Physics
2 answers:
IgorLugansk [536]4 years ago
3 0

I think it is B hope this helps

God Bless

Reika [66]4 years ago
3 0
Both rods and cones are sensitive to light. The difference between them is that the rods allow us to see<span> in very dim light but don't permit detection of </span>color<span>, while the cones let us </span>see color<span> but they don't work in dim light. When it gets dark the cones lose their ability to respond to light</span>
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A metal detector used in airports is actually a large coil of wire carrying a small current. Explain how it detects a gun, even
Alexus [3.1K]

<u>Metal detectors work by transmitting an electromagnetic field from the search coil into the ground. Any metal objects (targets) within the electromagnetic field will become energised and retransmit an electromagnetic field of their own. The detector’s search coil receives the retransmitted field and alerts the user by producing a target response. metal detectors are capable of discriminating between different target types and can be set to ignore unwanted targets. </u>

1. Search Coil

The detector’s search coil transmits the electromagnetic field into the ground and receives the return electromagnetic field from a target.

2. Transmit Electromagnetic Field (visual representation only - blue)

The transmit electromagnetic field energises targets to enable them to be detected.

3. Target

A target is any metal object that can be detected by a metal detector. In this example, the detected target is treasure, which is a good (accepted) target.

<em>hope this helps PLEASE MARK AS BRAINLIEST:)</em>

8 0
3 years ago
Runaway truck ramps are common on mountainous highways in case the brakes fail on large trucks. If a
dusya [7]

Answer:

W=-21,870,000\ J

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

W=\Delta k=k_1-k_0

Since

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}

W=0-21870000\ J

\boxed{W=-21,870,000\ J}

3 0
4 years ago
What metal do they use inside the torch to conduct electricity
sammy [17]
Inside the torch, they usually use copper or brass to conduct electricity.
7 0
3 years ago
What is the speed of a walking person in m/s if the person travels 2000 m in 35 minutes?
iren [92.7K]

Distance = 2000m

Time = 35 minutes = 35×60 sec = 2100 sec

We know, Speed = Distance/Time

Therefore, Speed

= 2000m/2100s = 20/21 m/s

Answer: 20/21 m/s

Hope it helps! Please do comment

4 0
3 years ago
A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling
Veronika [31]

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car \Delta m_ {s} , this happens during a time interval

[t, t + \Delta t], in addition to freight car and sand at time t.

In this way we need to define the two states:

State 1,

consider t, m_ {c} (t) + \Delta m_ {s} and V.

State 2,

consider t + \Delta t, m_ {c} (t), V + V \Delta V

In this state is the mass of sand output, which

is composed of

\Delta m_{s}, V + \Delta V

In this way we define the Linear movement in x, like this:

p_ {x} (t) = (\Delta m_ {s} + m_ {c} (t)) v

p_ {x} (t+\Delta t) = (\Delta m_ {s} + m_ {c} (t)) (v + \Delta v)

m_ {c} (t) = m_ {c, 0} - bt = m_ {c} + m_ {s} -bt

In this way we proceed to obtain the Force

F =\lim_{\Delta t \rightarrow 0} \frac {p_x (t + \Delta t) -p_ {x} (t)} {\Delta t}

F = lim_{\Delta t \rightarrow 0} m_ {c} (t) \frac {\Delta v} {\Delta t} + lim_{\Delta t \rightarrow 0} m_ {s} (t) \frac {\Delta v} {\Delta T}

Since the mass of the second term becomes 0, the same term is eliminated, thus,

F = m_ {c} (t) \frac {dv} {dt}

\int\limit ^ {v (t)} _ {v = 0} dv = \int\limit^t_0 \frac{Fdt} {m_ {c} + m_ {s} -bt}

V (t) = - \frac {F} {b} ln (\frac {m_c + m_s-bt} {m_c + m_s})

3 0
4 years ago
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