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New answer posted
7 months agoContributor-Level 10
Area of OABS is A?
Area of SCD is A?
Distance = |A? | + |A? |
A? = (1/2) [13/3 + 1]4 = 32/3
A? = (1/2) * 5/3 * 2 = 5/3
Distance = |A? | + |A? |
= 32/3 + 5/3 = 37/3
New answer posted
7 months agoContributor-Level 10
Moment of inertia in case (i) is I?
Moment of inertia in case (ii) is I?
I? = 2MR²
I? = (3/2)MR²
T? = 2π√ (I? /Mgd); T? = 2π√ (I? /Mgd)
T? /T? = √ (I? /I? ) = √ (2MR² / (3/2)MR²) = 2/√3
New answer posted
7 months agoContributor-Level 10
√2gh = (2r²g/9η) (ρ_t - ρ)
⇒ h = (2/81) (r? g (ρ_t - ρ)²/η²)
⇒ h ∝ r?
After falling through h, the velocity be equal to terminal velocity.
New answer posted
7 months agoContributor-Level 10
Till input voltage reaches 4 V. No zener is in breakdown region. So V? = V? then now hen V? changes between 4 V to 6 V one zener with 4 V will breakdown are P.D. across this zener will become constant and remaining potential will drop acro resistance in series with 4 V zener.
Now current in circuit increases Abruptly and source must have an internal resistance due to which. Some potential will get drop across the source also so correct graph between V? and t will be
We have to assume some resistance in series with source.
New answer posted
7 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
7 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
7 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
7 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
7 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
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