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Thepotemich [5.8K]
3 years ago
6

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz​

Physics
1 answer:
Ronch [10]3 years ago
8 0

Answer:

This is because these metals are used for minting (making) coins.

You might be interested in
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to
goldenfox [79]

Answer:

17304 J

Explanation:

Complete statement of the question is :

In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .

Part A

How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?

Solution :

T = tension force in the tow rope = 350 N

L  = length of the incline surface = 120 m

W_{t}  = work done by tension force = ?

The tension force acts parallel to incline surface, hence work done by tension force is given as

W_{t} = T L\\W_{t} = (350) (120)\\W_{t} = 42000 J

h = height gained by the rider = 30 m

m = total mass of rider and tube = 84 kg

Potential energy gained is given as

U = mgh\\U = (84) (9.8) (30)\\U = 24696 J

Q  = Thermal energy created

Using conservation of energy

Q = W_{t} - U\\Q = 42000 - 24696\\Q = 17304 J

7 0
3 years ago
Bobby is riding his bike down the road at a speed of 10 miles per hour. Ahead of him, he sees another rider, moving in the same
vekshin1
They will be travelling slower than 10mph.
if they were travelling at the same speed then they would stay an equal distance apart.
if they were travelling fatser then they would be getting further away more quickly than Bobby is catching up.
maybe they are travelling at 5mph but I'd say it's a safer option to chose under 10mph
4 0
3 years ago
Read 2 more answers
Explain how work is related to energy
Dahasolnce [82]

Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work.


So they are both closely related to each other.

HOPE THIS HELPS

3 0
3 years ago
You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m
ki77a [65]

So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

\boxed{\sf{\bold{t = \frac{s}{v}}}}

With the following condition :

  • t = interval of the time (s)
  • s = shift or displacement (m)
  • v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

  • h = height or any other displacement at vertical line = 19.6 m
  • g = acceleration of the gravity = 9.8 m/s²
  • v = velocity = 343 m/s

What was asked :

  • \sf{\sum t} = ... s

Step by step :

  • Find the time when the object falls freely until it hits the water. Save value as \sf{\bold{t_1}}

\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}

\sf{t_1 = \sqrt{4}}

\sf{\bold{t_1 = 2 \: s}}

  • Find the time when the sound propagate through air. Save value as \sf{\bold{t_2}}

\sf{t_2 = \frac{h}{v}}

\sf{t_2 = \frac{19.6}{343}}

\sf{\bold{t_2 \approx 0.06 \: s}}

  • Find the total time \sf{\bold{\sum t}}

\sf{\sum t = t_1 + t_2}

\sf{\sum t \approx 2 + 0.06}

\boxed{\sf{\sum t \approx 2.06}}

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

3 0
2 years ago
] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0
2 years ago
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