Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question
5.6k

Questions

0

Discussions

31

Active Users

0

Followers

New answer posted

7 months ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 21 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

g_e = g - Rω²
g? = g - (2gh/R)


Now Rω² = 2gh/R
h = R²ω²/2g

New answer posted

7 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-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

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1850k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.