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The electrons diffuse into the GROUND and the ground becomes slightly more NEGATIVELY CHARGED. The difference is nearly IMPOSSIBLE to measure, however.

u = 0 m/s

V = 25 m/s

t = 10 m/s

a) a = ?

As v = u + a t

so, 25 = 0 + a * 10

so, a = 25/10 = 2.5 m/s^2

b) s = ? (for FIRST 10 s)

As s = ut + 1/2 a t^2

so, s = 0 + 1/2 * 2.5 * 10^2

so. s = 125 m

c) s = ? ( for next 10 s)

s = v * t as a = 0 m/s^2

so, s = 25*10

so, s = 250 m

Hnce, TOTAL DISTNCE = 125 + 250

= 375 m

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If area is SMALLER, the PRESSURE will be higher as it is inversely proportional to area.

On differentiating very WELL get

a= 8t m/s ^2
now PUT t=1s
we get
a= 8 (1) = 8m/s^2........

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In mechanical ENGINEERING, backlash, SOMETIMES called
or play, is a clearance or lost motion in a mechanism caused by gaps between the parts. ... It can be SEEN when the direction of movement is reversed and the
or lost motion is TAKEN up before the reversal of motion is complete

By USING the FORMULA v=u+at
we have v=0+4(3)
so v=12m/s

In which STD you are STUDYING?
I'm not aware of the MEANING STRAIGHT slope....
Sorry

His AVERAGE speed for the JOURNEY is 57 KM PER hour

A) The substance will SINK as the density of the substance is greater than the density of the WATER,i.e. 2.5g/cm³>1 g/cm³
B) According to the Archimedes principle the MASS of the water displaced will be same as the mass of the substance .

Why Do Straws Work?

Because it is really the atmosphere that is doing the pushing, the atmospheric pressure LIMITS how high water will go up a straw. At SEA level, the AIR pressure is ENOUGH to SUPPORT a column of water about thirty feet high.

Even if you could suck all the air out of a forty-foot straw, the water wouldn’t rise more than thirty feet. In space, where there isn’t any atmospheric pressure, a straw wouldn’t work at all.

ANGLE will be 0° ....
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The FASTER speed of the GAS particles in hot AIR cause the particles to be more spread out than in cool air. Thus, hot air has a lower density and WOULD TEND to rise.

The DISPLACEMENT from the ORIGIN is APPROX. 6.3m in South-West DIRECTION

The drift velocity will remain same because CURRENT doesn't depend on CROSS SECTION AREA of conductors.

Mass of bottle = 20g

Mass of water = 2gm

We KNOW density of water is 1g/cc

∴ Volume = 2 cc

Mass of OIL = 1.76

Volume = 2cc

Density = Mass / Volume

= 0.88

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Greater resultant = a + b = 14
Least resultant = a_ b = 4

At 60° ... use law of COSINE
ANS is 12.28


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What causes long sight (hypermetropia)?
The causes of long sight are usually hereditary (genetic). Long sight can OCCUR at any age but it TENDS to become more noticeable above the age of 40 years.

In rare cases, long sight is caused by other conditions such as diabetes, small eye syndrome (microphthalmia), cancers around the eye and problems with the blood vessels in the retina.

Many BABIES and very YOUNG children tend to be slightly long-sighted but usually grow out of this by about 3 years of age.

1. Measurement errors :

They OCCUR when the scale is read incorrectly either by overreading the scale or undereading the scale to GIVE a lower or a higher reading of the ACTUAL measurement.

2. Recording mistakes occur when the readings are noted down wrongly. That is writing correct readings in the wrong COLUMNS or even writing different readings from the ones taken.

This happens either when one forget some decimal points or even INTERCHANGES two digits.

It's sounds more as a PSYCHOLOGICAL question....
and we FEEL it as stress and our BODY reacts to it...the only thing is that now-a-days no one CARES about it.

We need to CALCULATE the Centre of the gravity of the Indian map made up of the steel.

Since, Indian Map does not have the GEOMETRICAL significance. I means that it does not have the Proper shape like any shape in Geometry. Thus, we need to considered it has irregular.

Now, By considering it as irregular, we can calculate the Centre of the gravity by the method of finding the centre of gravity in case of Irregular lamina.

Steps for it ⇒

1. First create the small three hole on three opposite sides of the Indian Map. Then, From first hole, tie the thread and suspend the map. When the map is suspended, then the line going starlight to the bottom of the map is marked.

2. Same method is applied in second hole and third hole. Ans two more lines are marked on the map.

3. Note the Point of the Intersection of the three lines. Where these POINTS meets, the centre of the gravity of the Map lie at theta place.

This method can be used to find the centre of the gravity of any irregular lamina or any irregular body.

Hope it HELPS. :-)

Scientific METHOD is an empirical method of knowledge acquisition, which has characterized the development of natural science since at least the 17th century, involving careful observation, which INCLUDES rigorous skepticism about what one observes, given that cognitive assumptionsabout how the WORLD works influence how one interprets a percept; formulating hypotheses, via induction, based on such observations; experimentaltesting and measurement of deductions drawn from the hypotheses; and REFINEMENT (or elimination) of the hypotheses based on the experimental findings. These are principles of the scientific method, as OPPOSED to a definitive series of steps applicable to all scientific enterprises.

By LOOKING at the TYPE of IMG FORMED

Least count of stopwatch=RANGE of time/total divisions.

in the GIVEN CONDITION, divisions are 10,and range is 5seconds.

so its least count=5/10=1/2second

It's ANSWER isome 14.167 METER / SECOND

Given conditions ⇒

Mass =100 g

=0.10 kg

Velocity (V)=1.4 m/s

RADIUS(r)=1 m

The Tension in the string at the lowest point will be DUE to the weight of the bob and circular motion of the bob.

Therefore,

Tension = Force of gravity + Centripetal Force

∴ Tension = mg + mv²/r

∴ T = 0.10 × 10 + 0.10 × (1.4)²/1

⇒ T = 1 + 0.196 N

∴ T = 1.196 N

Hence, the tension in the string at the Instant = 1.196 N.

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MASS of the scooter(m) = 150 kg.

Radius(r) = 30 m.

Velocity of the Scooter(v) = 36 km/hr.

= 36 × 5/18 = 10 m/s.

Using the FORMULA,

Force = mv²/r

⇒ Force = 150 × 10²/30

∴ F = 500 N.

HENCE, the Centripetal Force of magnitude 500 N is acting on the scooter.

Hope it helps.

The Initial MOMENTUM is 100kgm/s and Final Momentum is 300kgm/s.... The Force ACTING on object is 20N....( I have considered time as 10s, as there's an ERROR in your QUESTION)

Answer:

Time taken = 1 hr 40 MIN 48 SEC = 1 hr4054min=17551hrs=75126hrs.

Let the ACTUAL speed be x km/hr.

Then, 75x×75126=42

⇒x=(5×12642×7×75)35km/hr.

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Acceleration DUE to the GRAVITY on Earth = 9.8 m/s².

We KNOW that the acceleration due to the gravity on the MOON is 1/6th that of the Earth.

∴ Acceleration due to the gravity on moon = 1/6 × 9.8 m/s².

= 1.63 m/s².

Hence, the acceleration due to the gravity on the moon is 1.63 m/s².

Hope it helps.

When there is ATOMIC DIFFUSION the all THREE particles ALPHA ,BETA,gamma are emitted with a high radiation

In PARALLEL and in SERIES

Your QUESTION is INCOMPLETE

1/8 -1/U=1/3 u is 24/5

Mass of the each Spherical BALLS = 10 kg.

Distance between the Spherical balls = 10 cm. = 0.1 m.

USING the FORMULA,

F = Gm₁m₂/r²

∴ F = 6.67 × 10⁻¹¹ × 10 × 10/(0.1)²

∴ F = 6.67 × 10⁻¹¹ × 10⁴

∴ F = 6.67 × 10⁻⁷ N.

∴ F ≈ 10⁴ G

Hence, the gravitational force of the attraction between the spherical balls is 6.67 × 10⁻⁷ N or 10⁴G N.

Hope it HELPS.

I have to GO to work and U for a few days but for now I'm going for me which for me and my family is going on with me and my wife would like to see the car and the other ONE is in for the night but I don't think I can do it for you if you want to come by and see you.

Average speed is 40kmph and total distance is 100km
so time taken is 100/40=2.5 hr
distance COVERED in 1st hris 30km
REMAINING 70KM should be COVERD in1.5 hr
so speed is 70/1.5=700/15=160/3=53.3kmph

dy/dt= [-SIN(5-3t)/5-3t]3

dy/dt= [-sin(5-3t)/5-3t]3dy/dt=-3sin(5-3t)/5-3t

dy/dt= [-sin(5-3t)/5-3t]3dy/dt=-3sin(5-3t)/5-3tanswer

Water has REFRACTIVE INDEX more than water. It DIVERGES the light BEAM and THUS

HOPE it HELPS................

Let the length of the bus be L m.

TIME TAKEN by the man to cross the bus is 18s

so, velocity of the man = L/18 m/s

Now time taken by the bus to cross the pole

is 4 s

so, velocity of the bus = L/4 m/s

Ratio = speed of the bus : the speed of the man

= L/4 : L/18

= 18 : 4

= 9 : 2

Given conditions ⇒

Initial Speed of the ball(u) = -5 m/s.

[Since, ball is dropped from the BALLOON].

Height of the balloon = 60 m.

Acceleration DUE to GRAVITY = 9.8 m/s²

Using the Formula,

S = ut + 1/2at²

⇒ 60 = -5t + 1/2 × 9.8 × t²

⇒ 60 = -5t + 4.9t²

⇒ 4.9t² - 5t - 60 = 0

∴ 49t² - 50t - 600 = 0

This is TH quadratic equation. If we solve it by the methods of the quadratic Formula,

we wlil get, t = - 4.02 and t = 3.03 seconds.

Since, time cannot be negative. Hence time will be 3 seconds.

Hence, the time taken by the ball is 3.02 seconds.

Hope it helps.

Aditya - L1 First Indian mission to study the Sun

The Aditya-1 mission was conceived as a 400kg class satellite carrying one payload, the Visible Emission Line Coronagraph (VELC) and was planned to launch in a 800 km low EARTH orbit.  A Satellite PLACED in the halo orbit around the LAGRANGIAN point 1 (L1) of the Sun-Earth system has the major advantage of continuously viewing the Sun without any occultation/ eclipses.  Therefore, the Aditya-1 mission has now been revised to “Aditya-L1 mission” and will be inserted in a halo orbit around the L1, which is 1.5 million km from the Earth.  The satellite carries additional six payloads with enhanced science scope and objectives.

The project is approved and the satellite will be launched during 2019 – 2020 timeframe by PSLV-XL from Sriharikota.

Aditya-1 was meant to observe only the solar corona.  The outer layers of the Sun, extending to thousands of km above the disc (photosphere) is termed as the corona.  It has a temperature of more than a million DEGREE Kelvin which is much higher than the solar disc temperature of around 6000K. How the corona gets heated to such high temperatures is still an unanswered question in solar physics. 

 Aditya-L1 with additional experiments can now provide observations of Sun's Photosphere (soft and hard X-ray), Chromosphere (UV) and corona (Visible and NIR).  In addition, particle payloads will study the particle flux emanating from the Sun and reaching the L1 orbit, and the magnetometer payload will measure the variation in magnetic field strength at the halo orbit around L1.   These payloads have to be placed outside the interference from the Earth’s magnetic field and could not have been useful in the low earth orbit.

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