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1 year ago

Please solve the Problem.

Physics

1 answer:

andrey2020 [161]1 year ago

8 0

The frequency heard by the moth is the frequency emitted by the bat by virtue of its speed; 38.0kHz. B) If the speed of the bat is increased, the frequency heard by the moth is higher.

<h3>What is the frequency heard by the moth?</h3>

It follows from the task content that the frequency of the emitted by the bat is heard by the moth and hence, the frequency heard by the moth is; 38kHz.

B) Since, the speed and frequency of sound waves are directly proportional, speed = wavelength × frequency; it therefore follows that when the speed of the bat is increased, the frequency heard by the moth is higher.

Read more on speed and frequency;

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- Find the velocity of a wave in the ocean if the wavelength is 2.0m and the frequency is 0.56Hz.

pentagon [3]

1.12 m/s is the velocity. You can get the velocity of a wave by multiplying the frequency and wavelength together. The product is the velocity.

6 0

3 years ago

A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

4 0

3 years ago

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

yuradex [85]

Since

Electric potential energy = qV

Where V = Ed

Hence
Electric potential energy = q(Ed) --- (1)

Since E = 1.0 * 10^3 N/C
d = 0.10 m
q = 4 * 10^-6 C

Plug in the values in (1)
(1) => Electric potential energy = 4 * 10^-6(1.0 * 10^3 * 0.10)
Electric potential energy = 400 μJ

3 0

3 years ago

Help with 7,8,9 please!!

Marysya12 [62]

# 7 the answer is 4) because you need opposites in order to attract

# 8 I'm going to guess it is 4) because electricity is needed in order a light bolt to work.(don't take my word for this one I'm not sure but, the other answer are correct)

# 9 the answer is 4) because the comb dose collect negative charges from the hair in order to make the hair have positive charge.

Hope this helped : )

4 0

3 years ago

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0

3 years ago

Read 2 more answers