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New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

  a = v 2 R

a = v 2 R c o s θ i ^ v 2 R s i n θ j ^

 

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  C P C v = y = 1 + 2 / f

 

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Heat Released by block = Heat gain by large Ice block

  m c Δ T = M i c e L              

->5 * 0.39 * 500 = mice * 335

  m i c e = 5 * 0 . 3 9 * 5 0 0 3 3 5              

= 2.91 kg

 

New answer posted

11 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Given

6a2 = 24

a2 = 4

 

a = 2m

Δ v = v ( γ ) Δ T                

= 3 α v Δ T              

= 3 * 5 * 1 0 4 * ( a 3 ) * 1 0                

= 3 * 5 * 1 0 4 * 8 * 1 0              

= 1 2 0 * 1 0 3 = 1 . 2 * 1 0 5 c m 3                

New answer posted

11 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

F. B. D of Beaker

 

By NLM2

f = m a = m ω 2 R m ω 2 R μ N R μ g / ω 2                              

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

X P ( t ) = α t + β t 2  

  V P ( t ) = α + 2 β t                            -(i)

X Q ( t ) = f t t 2               

V Q ( t ) = f 2 t                                 -(ii)

                             (i) = (ii)

α + 2 β t = f 2 t t ( 2 β + 2 ) = f α              

t = f α 2 ( β + 1 )         &

...more

New answer posted

11 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

First angle, θ1 = θ, R = u 2 s i n 2 θ g   

Another angle (θ2 = 90 - θ) for which range will be

Same as that of θ1 = θ

R 1 = u 2 s i n 2 ( 9 0 θ ) g = u 2 g s i n 2 θ = R

at  θ1=θ,h1=u2sin2θ2g

& θ 2 = 9 0 θ , h 2 = u 2 s i n 2 θ 2 2 g = u 2 c o s 2 θ 2 g

h 1 h 2 = 1 4 2 [ u 2 s i n 2 θ g ] 2  

R = 4 h 1 h 2  

So, Both statement is true & Reason is correct explanation for statement 1.

New answer posted

11 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

F=5i^+3j^7k^

r=2i^+2j^+k^

τ=|r*F|=|i^j^k^21537|=i^ (k13)j^ (145)+k^ (610)=17i^+19j^4k^

New answer posted

11 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

ΔKKi*100=KfKiKi*100

=Pf22mPi22mPi22m*100=Pf2Pi2Pi2*100

= (1.2Pi)2 (Pi)2Pi2*100=1.44Pi2Pi2Pi2*100

= 44%

New answer posted

11 months ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

m1 = 5kg

m2 = 3kg

As m1 is in rest

T = mgg = 30

N = m1 g cos

T = m1gsinθ50sinθ=30

sin =35θ=37°

N=50cosθ=50cos37°=50*45=40N

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