PHYSICS COLLEGE

If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. This changes the mass that oscillates (more water means more mass) but not the restoring force, which is determined by the stiffness of the glass itself. If you need to raise the frequency of a par- ticular glass, should you add water or remove water?

Answers

Answer 1

Answer:

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

$f=(1)/(2\pi )\sqrt{(k)/(m)}$

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency

Related Questions

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit. Required:Find the radius and period of the orbit.
A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?
A race car travels on a circular track at an average rate of 125 mi/h. The radius of the track is 0.320 miles. What is the centripetal acceleration of the car? 391 mi/h2 40.0 mi/h2 5,000 mi/h2 48,800 mi/h2
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by
The height h (in feet) of an object shot into the air from a tall building is given by the function h(t) = 650 + 80t − 16t2, where t is the time elapsed in seconds. (a) Write a formula for the velocity of the object as a function of time t.

If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?

Answers

Answer:

$heigth=2.86m$

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

$V_(avg)=(distance)/(time)\nV_(avg)=(1.6m)/(0.19s)\nV_(avg)=8.42m/s$

Also we know that average velocity

$V_(avg)=(V_(i)+V_(f))/2\n$

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

$V_(f)=V_(i)+gt\nV_(f)=V_(i)+(9.8)(0.19)\nV_(f)=V_(i)+1.862$

Substitute value of Vf in average velocity

$V_(avg)=(V_(i)+V_(f))/2\nwhere\nV_(f)=V_(i)+1.862\nand\nV_(avg)=8.42m/s\nSo\n8.42m/s=(V_(i)+V_(i)+1.862)/2\n2V_(i)+1.862=16.84\nV_(i)=(16.84-1.862)/2\nV_(i)=7.489m/s\n$

Vi is speed of balloon at top of the window

Now we need to find time

$V_(i)=gt\nt=V_(i)/g\nt=7.489/9.8\nt=0.764s$

So the distance can be found as

$distance=(1/2)gt^(2)\n distance=(1/2)(9.8)(0.764)^(2)\n distance=2.86m$

A person exerts a horizontal force of F=45N on the end of an 86cm wide door. The magnitude of the torque due to F about the pivot point is determined by |τ|=|rxF|=rFsinθ . Determine the magnitude of the torque, |τ| , on the door about its hinges due to F . |τ|=0Nm |τ|=38.7Nm |τ|=3870Nm

Answers

Answer: The magnitude of torque is 38.7Nm

Explanation: Please see the attachment below

The magnitude of the torque on the door about its h1nges due to the applied force is 38.7 Nm.

How to calculate the magnitude of the torque?

The magnitude of the torque on the door about its h1nges due to the applied force is calculated by applying the following formula as shown below;

τ = rF

where;

r is the perpendicular distance of the applied force
F is the applied force

The given parameters include;

perpendicular distance, r = 86 cm = 0.86 m

the applied force , F = 45 N

The magnitude of the torque on the door about its h1nges due to the applied force is calculated as;

τ = rF

τ = 0.86 m x 45 N

τ = 38.7 Nm

Learn more about torque here: brainly.com/question/30338159

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Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.

Answers

Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

Answers

Answer:

μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

a = 2 x / t²

a = 2 0.5 /0.5²

a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

X axis

fr - Wₓ = m a (1)

Y Axis

N- $W_(y)$ = 0

N = W_{y}

We use trigonometry to find the components of the weight

sin 45 = Wₓ / W

cos 45 = W_{y} / W

Wₓ = W sin 45

W_{y} = W cos 45

The out of touch has the expression

fr = μ N

fr = μ W_{y}

We substitute in 1

μ mg cos 45 - mg sin 45 = m a

W_{y} = (a + g sin 45) / g cos 45

μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

a = -4 m / s²

μ = 1- 4 / (9.8 cos 45)

μ = 0.423

Answer:

The μ = 0.422

Explanation:

The distance travelled by the mass is equal to:

$L=ut+(1)/(2)at^(2) \n0.5=(0*5)+(1)/(2) a(0.5^(2) )\na=4m/s^(2)$

The sum of forces in y-direction equals zero:

∑Fy = 0

N - (m * g * cosθ) = 0

N - (1 * 9.8 * cos45) = 0

N = 6.93 N

The sum of forces in x-direction is equal to:

∑Fx = ma

(m * g * sinθ) - fk = m * a

(1 * 9.8 * sin45) - fk = 1 * 4

fk = 2.93 N

fk = μ * N

2.93 = μ * 6.93

μ = 0.422

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A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 600lb, what is the size of the force when it is at each of the following distances from the Earths surface?a. 20,000 milesb. 30,000 milesc. 100,000 miles

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:

A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?