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solution: a body projected with velocity (u)=49m/s projected with an angle =60 degrees H=(u2sin2Ɵ)/2g H=(492sin260)/2g H=(49*49*3/4)/2*9.8 H=91.875m
solution: velocity of bullet after passing from plank ,its velocity decreases . initial velocity(u)=u final velocity(v)=initial velocity-1/20 of initial velocity v=u-u/20 v=19u/20 we have formula for one plank, (v)2-(u)2=2as, (19u/20)2-(u)2=2as 39u2/400=2as ……….(1) for n planks: then width of one plank will be ‘s’ width of n planks will be ns after passing from n planks …
రాష్ట్రంలో ప్రభుత్వ ఉపాధ్యాయులు పదోన్నతి పొందాలంటే టెట్ ఉతీర్ణులు కావడం తప్పనిసరి సర్కారు తుది నిర్ణయానికి వచ్చింది.
WE WILL GET DSC NOTIFICATION
solution: a car initially starts at rest (u)=0 it is moving with uniform acceleration and the car gains velocity(v)=20m/s. the time taken by the car is (t)=4s we have the formula v=u+ at 20=0+4a 20=4a a=5m/s2 we got the uniform acceleration a =5m/s2 then ,for displacement S=ut+1/2at2 =0*4+1/2*5*4*4 =1/2*5*2*4 =40m or displacement=(u+v)/2*t =(0+20)/2*4 =40m The …
answer: if a passenger tosses the coin, the coin will go up and it has only gravitational acceleration. but train moves with uniform acceleration towards. so, the coin will fall behind the passenger.
solution: The car starts from rest (u)=0 acceleration (a)=α to reach maximum velocity “V1“ then ,the car decelerates with (a)=-β final velocity (V2)=0 for the first case ,V1=0+αt1,t1=V1/α for the second case, initial velocity(u)=V1 with deceleration (a)=-β final velocityV2 =0 0=V1-βt2 V1=βt2,t2=V1/β Total time=V1/α+V1/β=V1(β+α)/αβ=V1=(αβ/α+β)t V1=(αβ/α+β)t
solution: For a freely falling body initial velocity (u)=0 for a body which travelled in nth second distance Sn=u+ a(n-1/2) for freely falling body acceleration(a)=g for 2nd second S2=g(2-1/2) =g(3/2) for 3rd second S3=g(3-1/2) =g(5/2) the ratio between is 3/2g:5/2g=3:5
solution: The object projected vertically upwards an initial velocity(u)=40m/s if the object to reach the maximum height, the final velocity will be zero. from kinematic equations ,we have the formula, (v)2-(u)2=2as here , v=0,u=40m/s, a=-g=10m/s2 0-40*40=2(-10)S -1600=-20S S=80m here, distance is nothing but maximum height .
