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speed of train is 72km/h = 72*5/18 = 20m/snow, time taken is 5 min = 300sec

so, a = V-u/t = 20/300 1/15 m/s²

now distance travelled iss = 1/2*a*t² = 1/2*(1/15)*(300)² = = 300*300/30 = 3000m

work done W = - delta UW = -[GMm(1/r1 - 1/r2)]W = -[GMm (1/R - 1/3R)]W = -GMm(2/3R)W = -2GMm / 3R joule

When lightning and thunder take place they occur together but the thunder is observed a little after the lightning.

Salt: Sulphate, Chloride or hydrogen Carbonate of Calcium or Magnesium (any one)

By adding equal amount of soap to water taken in two test tubes, shakingvigorously and comparing the length of lather or foam formed in each testtube.

note: I have assumed g to be 10 so consider little variation from actual answer. as most nearest answer is 44.1 so answer is a option

Here,

h = 2.7 m

a1 = 1.2 ms-2

g = 9.8 ms-2

u = 0 (w.r.t. elevator)

Net acceleration of the bolt, as the elevator is moving up with acceleration a,

Will be

a = a1 +g = 11 ms-2

Let time at which the ball starts dropping be t = 0

h = ½ at2

=> t = √(2h/a)

=> t = √(2×2.7/11)

=>t = 0.7 s

Now time when it reaches the floor from starting of the elevator will be = 2+ 0.7 = 2.7 s

Short Answer: It depends on the acceleration with which it is thrown downwards.

Explanation:The bubbles rise up due to the force of buoyancy which comes into play in fluids due to the increase in pressure as we go deeper into the fluid. The pressure, as we go a height h into the fluid is given by:

P=P(atmosphere) + hdG

Here,

P(atmosphere) is pressure due to the atmosphere which, on Earth, is equal to 10^5 Pascal.

h is the depth at which we are observing the pressure.

And G is theEFFECTIVEacceleration due to gravity.

Now if a is the vertical component of the acceleration vector,

G(effective) = g(vector) - a(vector)

g, as you must be knowing, is the acceleration due to gravity having a magnitude 9.8 m/s2 directed vertically downwards.

Thus the net buoyancy force, which is due to the pressure difference depends on G which in turn depends on the vertical acceleration you have given the fluid contained in the bottle.

Now, note this, since we are talking about vectors here, the downwards vertical is considered positive here, so,

If, a > g AND directed downwards,the bubbles will, wait for it, MOVE DOWNWARDS!

and if, a=g, as another guy already mentioned (:/) the bubbles will be motionless.

and if, a<g, the bubbles will move upwards as usual.

As we know g = GM/R²R= aso,g = GM/ (a)²= GM/ a²

so as the mass of the planet =2 Mdiameter = 2athan g = 2GM/ a^2

Given x the time interval is t,

let thr initial velocity be 0so, the equation of free fall is. s1 = (1/2)*gt² ..... (1)and velocity gained is V = gt

now for 2nd t interval the equation is

s2 = (gt)*t + (1/2)gt² so S2 = (3/2)gt² ............ (2)

now S2 -S1 = gt²(3/2 -1/2) = gt².

so, from here , we get t = √(S2-S1)/g

option 1.

Acceleration due to gravity. The body is Projectile.

Assume that to plot a graph of velocity and speedAccelerating rate is v/aDeccelerating rate is v/bTimd for acceleration t1and decceleration t2total time t sec is =t1+t2Time=velcity/rate(acc+decc)Here u need to find max velocity so,v/a+v/b=t1+t2v/a+v/b=tv=ab/a+b*t m/s

ksjsgsjao kskakskjs jsjsjiw

500gm10cm k=100N/mf=kxf=100×10F=1000

500gm10cmK=100n/mf=100×10F=1000

as it is from resthence u=0a=v-u/ta=8-0/2=4m/s^2henceforce required=ma=3*4=12N

Newton's kinematics eqns -- Newton's kinematic equations are used to describe relationship between distance, time, velocity and acceleration.- These are only applicable for uniformly accelerated motion.- Standard kinematics equations(1) v = u + at(2) s = ut + 1/2 at^2(3) v^2 = u^2 + 2as

◆ Kinematic equations for freely falling body-Here,Acceleration a = gDisplacement s = h,Initial velocity u = 0

(1) First kinematics eqn -v = gt(2) Second kinematics eqn -h = 1/2 gt^2(3) Third kinematics eqn -v^2 = 2gh

s=1500m she his right answer

The image below contains the graph of the motion of the body.

The area under the graph gives the distance.

The area is a trapezium.

Formulae for getting area of a trapezium :

(a+b) /2 × h

a=6sb=12sh=20m/s

A=(6+12)/2 ×20

9×20=180m

Retardation is -ve of acceleration

so, R = -(a) = -(-2.5)m/s² = 2.5m/s².

Retardationmeans that the director of change is opposite to the current velocity so it will decrease. Retardation eats the current velocity until it becomes 0 and then the body will start moving in opposite direction.The examples of retardation(decrease in velocity) are everywhere.

If there was no retardation then no body will ever be able to stop.

Example- as in daily life we walk, we thrown ball, we drive car and all of them eventually stop means their velocity decreases.

Acceleration of a body is defined as the rate of change of its velocity. So, when the velocity of a body is not uniform (it is changing), the motion of body is said to be accelerated.

Example:Suppose a car starts off from rest (initial velocity is zero) and its velocity increases at a steady rate so that after 5 sec its velocity is10 m/s ,then the car is said to have an acceleration of 10-0/5= 2m/s square and motion is said to be accelerated motion.

Retardation :

means DECELERATION

When the velocity of the body is decreasing ,it is said to beretardation.

When a body moves in a negative acceleration ,it is calledretardation.

EXAMPLE:

when we apply brake , the car decelerates , this is known as retardation

V = u+at , u = 180km/h = 180*5/18 = 50m/s

when car stops at t = 5 sec , then v = 0

so, 0 = 50 +a*5

=> a = -50/5 = -10 m/s²

avg retardation = 10m/s²

thanks

The car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s. The acceleration is found to be,

a= (v – u)/t = (8 – 0)/10 =0.8 m/s²

Distance traveled during this time is,

S = 0 + ½ at²

=> S = 0.5 X 0.8 X 10²

=> S = 40 m

Suppose it travels 'x' distance with constant velocity, 8 m/s, for time 't'.

It then travels 64 m with uniform retardation and comes to rest.

Total distance traveled = 40+x+64 = 584

=> x = 480 m

So, 8 X t = 480

=> t = 60 s

Let the car travel 64 m with uniform retardation for time t

Using,

v²= u²- 2aS

=> 0 = 8²- 2(a)(64)

=> a = 0.5 m/s²

Retardation = 0.5 m/s²

t= (8)/(0.5) = 16 s

Total time taken is= 10 s + 60 s + 16 s =86 s

v=0

t=10

a=-2.5

v=u+at

0=u+(-2.5+10)

u=25m/s

Since,v=u+atV=12.5U=16.6T=1012.5=16.66+a*1012.5-16.6=10a-4.1=10aa=-4.1/10=-0.41m/s^2So retardation=0.41m/s^2

Hence, option (c) is correct.

Thank you

The minimum time in which it can cover a total distance of 1.5km (= 1500m)is 30s

you accelerate from rest during a time t ==> your speed = 5tthen you decelerate with an initial speed = 5t and a final speed = 0during a time T0 = 5t - 10T ==> t = 2T

total distance must be 1500mdistance during acceleration = 5t²/2 = 10T²distance during deceleration = (5t)T - 10T²/2 = 10T²- 5T² = 5T²10T² + 5T² = 1500T² = 100T = 10s and t = 20stotal time = 30s ....ans

T=20min time hour =20×1/60=1/3D =15kms=d/t=15/1/3=45km/h time=20×60=1200sd=15×1000=15000ms=d/t=15000/1200=12.5m/s

500×0.5=250so 250 is the right answer....

250 is correct answer of following question

1.6× 10_¹9 is the right answer

The answer is 30s.

there is a formula for this type of questions - [(αβ)/2(α+β)] * t2 = s

where α = Maximum acceleration t = time

β = Maximum retardation s = distance

Solution: Let (D) represent the distance covered during acceleration.

Then (1500 - D) would be the distance covered during negative acceleration.

Let t (1) represent the time lapse during acceleration, and t (2) represent the time lapse during negative acceleration. The total time T would be [t (1) + t (2)]; and T would be the measurement of time that we are seeking.

The definition of acceleration is delta V / delta t. Since we are taking that it started from rest, then the velocity at t = t (1) would be: V (1) = 5 t (1).

Since we are taking it that the cycle came to rest at the end, V (1) = 10 t (2).

Substitution axiom tells us that 5 t (1) = 10 t (2). This reduces to t (2) = half t (1) which makes total sense since retardation is double the acceleration.

Using kinematic equation [d = V (i) t + 1 / 2 a t^2] for the first leg, we get:

D = 1/2 (5) [t (1)]^2 and for the second leg: (1500 - D) = [{5 t (1)} {t (2)}] + 1/2 (-10) [t (2)^2]

Substitution for t (2) gives us: (1500 - D) = [{5 t (1)} {t (1)/2}] + 1/2 (-10) [{ t (1)}/2]^2

Substitution for D gives us: [1500 - (1/2) (5) {t (1)}^2] = [{5 t (1)} {t (1)/2}] + 1/2 (-10) [{ t (1)}/2]^2

Remove brackets and solve for t (1). We get:

1500 - (5/2) [t (1)^2)] = 5/2 [t (1)^2] - (5/4) [t (1)^2]

6000 / 15 = t (1)^2 → t (1) = 20 seconds

Then t (2) = 20/2 = 10 seconds. Total trip = 30 seconds.

Given,m1 = 100 g = 0.1 kgm2 = 500 g = 0.5 kgm3 = 50 g = 0.05 kg

The free-body diagram for the system is shown below:

From the free-body diagram of the 500 g block,T + 0.5a − 0.5g = 0 .....i

From the free-body diagram of the 50 g block,T1 + 0.05g − 0.05a = a ....ii

From the free-body diagram of the 100 g block,T1 + 0.1a − T + 0.5g = 0 ....iii

From equation ii,T1 = 0.05g + 0.05a .....iv

From equation i,T1 = 0.5g − 0.5a .....v

Equation iii becomesT1 + 0.1a − T + 0.05g = 0

From equations iv and v, we get:0.05g + 0.05a + 0.1a − 0.5g + 0.5a + 0.05g = 00.65a = 0.4 g⇒a=0.40.65g =4065g=813g (downward)

So, the acceleration of the 500 gm block is 8g13downward.

Specific heat of water is 1 cal/g.

There is totally 1kg, so it possesses 1000cal of energy.

Now, since ice is already at 0 Celsius, it needs latent heat to turn into water.

This energy is provided by the water. Latent heat of ice is 80cal/g.

So, for 1kg, it requires 8000cal. But water only possess 1000cal of energy.

So, only certain mass of ice will get converted into water at 0C, and the rest of the water at 10C will cool down to 0C also.

But, since only 1000cal is available, part of the ice remains as ice.

So, temperature in the end is 0 Celsius.

A.

how plz explain full

when we mix ice in water that is 0° it melts to upto 0° and when this cold water is mixed with 10°hot water the cold water gets some heat from hot but not reach upto 10°