Isotope what is the meaning

Physics

1 answer:

Nesterboy [21]2 years ago

3 0

Isotope means that a chemical element that has the same number of protons but neutron number differs.

<u>Explanation:</u>

In isotope, the chemical element differs in neutron and nucleon number. Thus, different isotopes of a single component hold the same place in the periodic table.

Within the atom’s nucleus, protons are defined as an atomic number that is significantly equal to electrons in a neutral atom. An isotope of a given element has a different mass number. In general, every atomic number has a specific element, but in isotope, an atom may have a wide range of neutrons.

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A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of

Katen [24]

Answer:

S = V t where S is the horizontal distance traveled

1/2 g t^2 = H where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

Giving V in terms of L and theta

7 0

1 year ago

A car speeds over a hill past point A, as shown in the figure. What is the maximum speed the car can have at point A such that i

olchik [2.2K]

Answer:

see explanations below

Explanation:

At the point when the car leaves the track, the reaction on the road is zero, meaning that the centrifugal force equals the gravitation force, namely

mv^2/r = mg

Solve for v in SI units

v^2 = gr = 9.81 m/s^2 * 14.2 m = 139.302 m^2/s^2

v = sqrt(139.302) = 11.8 m/s

Answer: at 11.8 m/s (26.4 mph) car will leave the track.

3 0

2 years ago

I need to lift a 2000kg car, 1.798m and the joules required is 35240.8. Converted to watt (W = 35240.8/5 (s)) I got 7048.16 W. I

marusya05 [52]

This is a very interesting problem ... mainly because it's different from
the usual questions in the Physics neighborhood.

I can discuss it with you, but maybe not quite give you a final answer
with the information you've given in the question.

I agree with all of your calculations so far ... the total energy required,
and the power implied if the lift has to happen in 5 seconds.

First of all, let's talk about power. I'm assuming that your battery is
a "car" battery, and I'm guessing you measured the battery voltage
while the car was running. Turn off the car, and you're likely to read
something more like 13 to 13.8 volts.
But that's not important right now. What I'm looking for is the CURRENT
that your application would require, and then to look around and see whether
a car battery would be capable of delivering it.

   Power = (volts) x (current)

   7,050 W = (14 volts) x (current)

   Current = (7,050 watts / 14 volts) = 503 Amperes.

That kind of current knocks the wind out of me. I've never seen
that kind of number outside of a power distribution yard.
BUT ... I also know that the current demand from a car battery during
starting is enormous, so I'd better look around online and try to find out
what a car battery is actually capable of.

I picked a manufacturer's name that I'd heard of, then picked their
recommended battery for a monster 2003-model car, and looked at
the specs for the battery.

The spec I looked at was the 'CCA' ... cold cranking Amps.
That's the current the battery is guaranteed to deliver for 30 seconds,
at a temperature of 0°F, without dropping below 12 volts.

This battery that I saw is rated 803 Amps CCA !

OK. Let's back up a little bit. I'm pretty sure the battery you have
is a nominal "12-volt" battery. Let's say you use to start lifting the lift.
As the lift lifts, the battery voltage sags. What is the required current
if the battery immediately droops to 12V and stays there, while delivering
7,050 watts continuously ?

Power = (volts) x (current)

7,050 W = (12 V) x (current)

Current = (7,050 W / 12 V) = 588 Amps .

Amazingly, we may be in the ball park.
If the battery you have is rated by the manufacturer for 600 Amps
CCA (0°F) or CA (32°F), then the battery can deliver the current
you need.
BUT ... you can't conduct that kind of current through ear-bud wire,
or house wiring wire. I'm not even so sure of jumper-cables.
You need thick, no-nonsense cable, AND connections with a lot of
area ... No alligator clips. Shiny nuts and bolts with no crud on them.

Now ... I still want to check the matter of the total energy.
I'm sure you're OK, because the CCA and CA specifications talk about
30 seconds of cranking, and you're only talking about 5 seconds of lifting.
But I still want to see the total energy requirement compared to the typical
battery specification ... 'AH' ... ampere-hours.

You're talking about 35,000 joules

      = 35,000 watt-seconds

   = 35,000 volt-amp-seconds.

   (35,000 volt-amp-sec) x (1 hour/3600 sec) / (12 volt)

   = (35,000 x 1) / (3600 x 12) volt-amp-sec-hour / sec-volt

   =    0.81 Amp-Hour .

That's an absurdly small depletion from your car battery.
But just because it's only 810 mAh, don't get the idea that you can
do it with a few rechargeable AA batteries out of your camera.
You still need those 600 cranking amps. That would be a dead short
for a stack of camera batteries, and they would shrivel up and die.

Have I helped you at all ?

5 0

2 years ago

4.A 31-cm long conducting wire of 9-g carrying 7- A current is placed in a uniform magnetic field. What are the strength and dir

Anika [276]

Answer

Given,

Length of the wire,L = 31 cm = 0.31 m

mass of the wire, m = 9 g = 0.009 Kg

Current in the wire,I = 7 A

Magnetic field strength, B= ?

Equating magnetic force to the weight of the wire.

$BIL = m g$