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New answer posted
4 months agoContributor-Level 10
dm (t)/dt = bv²
F_thrust = v ( dm/dt )
Force on satellite = -v (dm (t)/dt)
M (t)a = -v (bv²)
a = - (bv³)/M (t)
New answer posted
4 months agoContributor-Level 10
Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3
New answer posted
4 months agoContributor-Level 10
f? = frequency heard by wall = f_s (v_s/ (v_s - v_c)
f? = frequency heard by driver after reflection from wall
f? = (v_s + v_c)/v_s)f? = (v_s + v_c)/ (v_s - v_c)f?
f? /f? = (v_s - v_c)/ (v_s + v_c)
48/44 = (v_s - v_c)/ (v_s + v_c)
12 (v_s + v_c) = 11 (v_s - v_c)
23v_c = v_s
v_c = v_s/23 = 345/23 = 15 m/s
= (15 * 18)/5 = 54 km/hr
New answer posted
4 months agoContributor-Level 10
Potential of centre, = V =
Vc = K (Σq)/R
Vc = K (0)/R = 0
Electric field at centre E_B = E_B = ΣE
Let E be electric field produced by each
charge at the centre, then resultant electric field will be
Ec = 0, since equal electric field vectors are acting at equal angle so their resultant is equal to zero.
New answer posted
4 months agoContributor-Level 10
Energies of given Radiation can have
The following relation
Eγ-Rays > EX-Rays > Emicrowave > EAM Radiowaves
∴ λγ-Rays < X-Rays < microwave < AM Radiowaves
According to tres.
(a) Microwave → 10? ³ m
(b) Gamma Rays → 10? ¹? m (ii)
(c) AM Radio wve → 100 m (i)
(d) X-Rays → 10? ¹? m
New answer posted
4 months agoContributor-Level 10
M = µ? NiA
Here
µ? = Relative permeability
N = Number of turns
i = Current
A = Area of cross section
M = µ? NiA = µ? nliA
M = µ? niV = 1000 (1000) (0.5) (10? ³) = 500 = 5 * 10²Am²
New answer posted
4 months agoContributor-Level 10
In adiabatic process
PV? = constant
P (m/ρ)? = constant
As mass is constant
P ∝ ρ?
P_f/P_i = (ρ_f/ρ_i)? = (32)? /? = 2? = 128
New answer posted
4 months agoContributor-Level 10
Now, using junction analysis
We can say, q? + q? + q? = 0
2 (x - 6) + 4 (x - 6) + 5 (x) = 0
x = 36/11, q? = 36 (5)/11 = 180/11
q? = 16.36µC
New answer posted
4 months agoContributor-Level 10
At T°C L = L? + L?
At T +? T Leq = L'? + L'?
where L'? = L? (1 + α? T)
L'? = L? (1 + α? T)
Leq = (L? + L? ) (1 + αavg? T)
⇒ (L? + L? ) (1 + αavg? T) = L? + L? + L? α? T + L? α? T
⇒ (L? + L? )αavg = L? α? + L? α?
⇒ αavg = (L? α? + L? α? )/ (L? + L? )
New answer posted
4 months agoContributor-Level 10
Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.
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