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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A set of 20 tuning forks is arranged in series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is ___________Hz.

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Vishal Baghel

Contributor-Level 10

A/c to question

f + 76 = 2f

f = 76

last frequency

= f + 76 = 2 * 76

= 152 H₂

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6 months ago

Physics Laws of Motion

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

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Payal Gupta

Contributor-Level 10

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h = u t + \frac{1}{2} a t^{2}$

$= 0 * 1 + \frac{1}{2} * 10 * 1 * 1$

= 5m

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A geyser heats water flowing at a rate of 2.0kg per minute from 30°C to 70°C. If geyser operates on a gas burner, the rate of combustion of fuel will be________g min^-1.

[Heat of combustion = 8 * 10³ Jg^-1,

Specific heat of water = 4.2Jg^-1 °C^-1]

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Vishal Baghel

Contributor-Level 10

Heat given to water by geyser

$H = {\overset{o}{m}}_{ω} C Δ T$

$= 2 * 1 0^{3} \frac{g r a m}{m i n} * 4.2 J g^{- 1} c^{- 1} * (70 - 30) ° C$

$= 2 * 42 * 4 * 1 0^{3} J / m i n$ - (1)

heat provided by combustion of fuel

$H = {\overset{o}{m}}_{f} C_{f}$

${\overset{o}{m}}_{f} = \frac{2 * 42 * 4}{8} = 42$

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A system to 10 balls each of mass 2kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth in figure. Tension on the string between the 4^th and 8^th ball is_________N when 6^th ball just leaves the table.

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Vishal Baghel

Contributor-Level 10

12 g – T = 12 a- (1)

T – T' = 2a- (2)

T' = 6a- (3)

From (1) & (2)

12g – 12a – T' = 2a- (4)

From (4) & (3)

12g – 12a – T' = 14

$12 g = \frac{10}{3} T' \Rightarrow T' = \frac{120 * 3}{10} = 36$

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6 months ago

Physics Work, Energy and Power

A fighter jet is flying horizontally at a certain with a speed of 200ms^-¹. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle q with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of q will be _____________.

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Payal Gupta

Contributor-Level 10

B fighter jet

A anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

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New answer posted

6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A batsman hits back a back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15ms^-1. The impulse imparted to the ball is__________Ns.

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Vishal Baghel

Contributor-Level 10

Impulse = change in momentum

= $Δ P = 2 m v$

= 2 * 15 * 0.4

= 12 N sec

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A ball is projected vertically upward with an initial velocity of 50ms^-1 at t = 0s. At t =2 2s, another ball is projected vertically upward with same velocity. At t = _______s, second ball will meet the first ball (g = 10 ms^-2).

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Vishal Baghel

Contributor-Level 10

$h = 50 t - \frac{1}{2} ? * 10 t^{2} = 80 m$

Now velocity B with respect to A

$V_{B A} = V_{B} - V_{A}$

= 50 – 30

= 20 m/sec

Now $A_{B A} = a_{B} - a_{A}$

= $10 (- \hat{j}) - (- 10 \hat{j})$

= 0

Time taken to collide.

$h = V_{B A} t$

80 = 20t

t = 4 sec

Total time = 4 + 2

= 6 sec

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6 months ago

Physics Communication Systems

Choose the correct statement for amplitude modulation

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Payal Gupta

Contributor-Level 10

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A sinusoidal wave y(t) = 40 sin (10 * 10⁶pt) is amplitude modulated by another sinusoidal wave x(t) = 20sin(1000πt). The amplitude of minimum frequency component of modulated signal is:

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Vishal Baghel

Contributor-Level 10

$C_{m} (t) = A_{C} s i n ω_{C} t + \frac{μ A_{C}}{2} [c o s (ω_{C} - ω_{m}) t - c o s (ω_{C} + ω_{m}) t]$

$μ = \frac{A m}{A_{C}}$

$A_{C} = 40$

$μ = \frac{1}{2}$

So amplitude of minimum frequency = $\frac{μ A_{C}}{2}$

= 10

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