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ivann1987 [24]
3 years ago
15

If you slosh the water back and forth in a bathtub at the correct frequency, the water rises first at one end and then at the ot

her. Suppose you can make a standing wave in a 157 cm long tub with a frequency of 0.3165 Hz. What is the velocity of the water wave? Answer in units of m/s.
Physics
1 answer:
notka56 [123]3 years ago
7 0

Answer:

Velocity(v) = frequency(f) × wavelength

f = 0.3165

Wavelength = 2×length(L)

L = 157cm

Convert the length in centimetres to metre = 1.57m

v = 2×1.57 × 0.3165

v = 0.99m/s

Approx. 1m/s

Explanation:

The velocity of a wave is the product of its frequency and it's wavelength. The frequency is already known. The wavelength is the distance between two successive wave crests which is formed by sloshing water back and forth in the bath tub. Sloshing water to one end of the tub will produce a wave crest first at that end then the other completing a cycle. The wavelength will be twice the length of the bath tub as it is the distance that both crests are formed.

Wave crest is the highest point of a wave, and in this case is where the water rises to a high point in the bath tub

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How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav
JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

so maximum height on moon is 6 times more than the maximum height on Earth

8 0
2 years ago
A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

4 0
3 years ago
What are the 3 basic parts of an atom and what are their charges?
kotegsom [21]

Answer:

electrons neutrons and protons

Explanation:

electron are negative

nuetron are neutral

protons are positive

7 0
3 years ago
A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli
AleksandrR [38]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s

Time taken by the train to stop is 20 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m

∴ The engineer applied the brakes 500 m from the station

4 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
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