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3 years ago

What forces keep an air hockey puck hovering motionless above an air hockey table?

Physics

2 answers:

Mnenie [13.5K]3 years ago

4 0

<span>C. the puck's weight and the air blowing </span>

Ksenya-84 [330]3 years ago

4 0

Answer:

C. the puck's weight and the air blowing

Explanation:

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Quiz Review Problems

zlopas [31]

The momentum of a neutron p = 586.25 kg m / s.

<u>Explanation:</u>

The product of mass and the velocity gives the momentum of an object and it is a vector quantity. It is denoted by the letter p. The unit of momentum is kilogram meter per second (or) kg m / s.

Given mass m = 1.675 $\times$ 10, velocity v = 3.500 $\times$ 10

Momentum, p = mv

where m represents the mass,

v represents the velocity.

momentum p = (1.675 $\times$ 10) $\times$ (3.500 $\times$ 10)

momentum p = 586.25 kg m / s.

5 0

3 years ago

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo

S_A_V [24]

Answer:

A. Z = 185.87Ω

B. I = 0.16A

C. V = 1mV

D. VL = 68.8V

E. Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

$Z=\sqrt{R^2+\omega^2L^2}$ (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

$Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega$

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

$I=\frac{V}{Z}$ (2)

V: voltage amplitude = 30.0V

$I=\frac{30.0V}{185.87\Omega}=0.16A$

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

$V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV$ (3)

D. The voltage across the inductor is:

$V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V$

E. The phase difference is given by:

$\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°$

5 0

3 years ago

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The

Rufina [12.5K]

Answer:

<h2>4.25m/s</h2><h2>E. None of the option is correct</h2>

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

4 0

4 years ago

A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motio

QveST [7]

Answer:

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ = 60/(25×9.8)

μ = 60/245

μ = 0.245

3 0

3 years ago

Label these parts on the wave below: Amplitude, Wavelength, Crest, Trough, Rest Position

Leviafan [203]

Answer:

Wavelength is the distance between from one crest to another crest or from one trough to another trough. The amplitude is the distance from the midpoint to the crest or trough. Crest is the highest point of the or a wave. Tough is the lowest point of the or a wave. Rest position is the position where it lies on the midpoint line.

Explanation:

I need a diagram to label these parts.

5 0

3 years ago