A small airplane flies 9000 moters east in 60 seconds. What is the airplane's velocity?

Consider a taut inextensible string. You shake the end of the string with some frequency, causing a wave to travel down the stri

masha68 [24]

Answer:

Part 1:

Option B is correct (It will remain unchanged).

Part 2:

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Option E is correct (It will be half as fast/long.)

Part 4:

Option C is correct (It will increase by a factor of √ 2.)

Explanation:

Formula we are going to use:

V=f*λ

Where:

V is the speed of Sound

f is the frequency of wave

λ is the wavelength.

The speed of wave , tension and linear density have following relation:

$V=\sqrt{F/\rho}$

Where:

V is the speed of Sound (Initial)

F is the tension in string (Initial)

$\rho$ is the linear density of string (Constant)

Terms:

V' is the new speed

f' is the new frequency

λ' is the wavelength

Solution:

Part 1:

From $V=\sqrt{F/\rho}$ :

Speed of Sound is independent of the frequency of shaking so speed well remain unchanged.

Option B is correct (It will remain unchanged)

Part 2:

If F'=2F then

$V=\sqrt{F/\rho}$

$V'=\sqrt{F'/\rho}\\V'=\sqrt{2F/\rho}\\V'= \sqrt{2} * \sqrt{F/\rho}\\V'=\sqrt{2}V$

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Formula we are going to use:

V=f*λ

Given f'=2f,

Even though frequency is doubled we will keep velocities same. V=V' in order to find the changing wavelength.

V'=f'*λ'

f*λ=f'*λ'

f*λ=2f*λ'

Solving above Equation:

λ'=λ/2

Option E is correct (It will be half as fast/long.)

Part 4:

T'=2T means $V'=\sqrt{2}V$ (From Part 1)

f'=f

Now:

V'=f'*λ'

$\sqrt{2}f*\lambda=f'*\lambda '\\\sqrt{2}f*\lambda=f*\lambda '\\ \lambda '=\sqrt{2}*\lambda$

Option C is correct (It will increase by a factor of √ 2.)

5 0

3 years ago

A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?

ladessa [460]

The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N

7 0

3 years ago

When using the magnification equation, a value greater than 1 as the solution for M indicates that the image is_____________.

Alexxandr [17]

Larger than the object.

6 0

4 years ago

Read 2 more answers

A 19.0-g bullet embeds itself in a 8.0-kg wooden block, which rests on a horizontal frictionless surface and is attached to a ho

Anton [14]

Answer:

d) 700 m/s

Explanation:

if k is the force constant and x is the maximum compression distance, then:

the potential energy the spring can acquire is given by:

U = 1/2×k×(x^2)

and, the kinetic energy system is given by:

K = 1/2×m×(v^2)

if Ki is the initial kinetic energy of the system, Ui is the initial kinetic energy of the system and Kf and Uf are final kinetic and potential energy respectively then, According to energy conservation:

initial energy = final energy

Ki +Ui = Kf +Uf

Ui = 0 J and Kf = 0J

Ki = Uf

1/2×m×(v^2) = 1/2×k×(x^2)

m×(v^2) = k×(x^2)

v^2 = k×(x^2)/m

= (500)×((21×10^-2)^2)/(19×10^-3 + 8)

= 2.75

v = 1.66 m/s

the v is the final velocity of the bullet block system, if m1 is the mass of bullet and M is the mass of the block and v1 is the initial velocity of the bullet while V is the initial velocity of the block, then by conservation linear momentum:

m1×v1 + M×V = v×(m1 + M) but V = 0 because the block is stationary, initially.

m1×v1 = v×(m1 + M)

v1 = v×(m + M)/(m1)

= (1.66)×(19×10^-3 + 8)/(19×10^-3)

= 699.86 m/s

≈ 700 m/s

Therefore, the velocity of the bullet just before it hits the block is 700 m/s.

5 0

3 years ago

Which of the following is the most likely end for a star that formed from a very small nebular cloud? . . Black hole . . Planeta

vekshin1

The answer for this question would be Planetary Nebula.

Black holes are created when the star core has a mass of more than 2.5 times of the Sun. In Supernova, for stars with mass of more than 8 times the mass of the Sun, death is signalled by a gigantic explosion: during the first second it can be as bright as a whole galaxy with hundreds of billions of stars. In Red Giants, it is due to explosion of average stars like the Sun. Lastly, in Planetary Nebula, for small stars (that is less than 8 times the mass of the Sun), at the end of the Red Giant phase, the star can’t contract enough to generate the temperatures needed for further nuclear fusion.

8 0

3 years ago