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4vir4ik [10]
3 years ago
7

If the wavelength of a sound wave increases and the frequency of the sound wave does not change what happens to the speed of the

wave?
a. it will stay the same

b. it will decrease

c. it cannot be determined

d. it will increase
Physics
1 answer:
salantis [7]3 years ago
3 0

If the wavelength of a sound wave increases and the frequency of the sound wave does not change, the speed of the wave will increase.

Ans: D

Explanation

The sound wave speed is given by E=fλ, where f indicates its frequency and λ indicates its wavelength.

From the equation, it is evident that the sound speed is proportional to both frequency and wavelength.

Here, as wavelength increases, wave speed increases provided there is no change in frequency.

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A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of
Nonamiya [84]

Answer: Impulse = 4 kgm/s

Explanation:

From the question, you're given the following parameters:

Momentum P1 = 12 kgm/s

Momentum P2 = 16 kgm/s

Time t = 0.2 s

According to second law of motion,

Force F = change in momentum ÷ time

That is

F = (P2 - P1)/t

Cross multiply

Ft = P2 - P1

Where Ft = impulse

Substitute P1 and P2 into the formula

Impulse = 16 - 12 = 4 kgm/s

The magnitude of the impulse is therefore 4 kgm/s.

6 0
3 years ago
A torsional pendulum consists of a disk of mass 450 g and radius 3.5 cm, hanging from a wire. If the disk is given an initial an
Montano1993 [528]

To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:

\omega = 2\pi f

\omega = 2\pi (2.5)

\omega = 5\pi rad/s

The angular displacement is given as the form:

\theta (t) = \theta_0 cos(\omega t)

In the equlibrium we have to t=0, \theta(t) = \theta_0 and in the given position we have to

\theta(t) = \theta_0 cos(5\pi t)

Derived the expression we will have the equivalent to angular velocity

\frac{d\theta}{dt} = 2.7rad/s

Replacing,

\theta_0(sin(5\pi t))5\pi = 2.7

Finally

\theta_0 = \frac{2.7}{5\pi}rad = 9.848\°

Therefore the maximum angular displacement is 9.848°

6 0
3 years ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
Pachacha [2.7K]

Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

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By comparing we see that a = g/2.

5 0
3 years ago
The gravitational force between two objects depends on the masses and what factor between them
mina [271]
Mass and distance

If mass is doubled, the force of gravity between the objects is doubled
6 0
3 years ago
A diode vacuum tube consists of a cathode and an anode spaced 5-mm apart. If 300 V are applied across the plates. What is the ve
MAXImum [283]

Answer:

Explanation:

There is electric field between the plates whose value is given by the following expression

electric field E =  V /d where V is potential between the plates and d is distance between them

E = 300 / 5 x 10⁻³

=  60 x 10³ N/c

Force on electron = q E where q is charge on the electron

F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.

Acceleration a = force / mass

a = 96 x 10⁻¹⁶/ mass  = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹

= 10.55 x 10¹⁵ m / s²

For midway , distance travelled

s =  2.5 x 10⁻³ m

s      =  1\2 a t²

t = \sqrt{\frac{2s}{a\\ } }

= \sqrt{\frac{2\times2.5\times10^{-3}}{ 10.55\times10^{15}}

t = .474 x 10⁻¹⁸ s

For striking the plate time is calculated as follows

t = [tex]\sqrt{\frac{2\times5\times10^{-3}}{ 10.55\times10^{15}}[/tex]

t = 0.67 x 10⁻¹⁸ s

3 0
3 years ago
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