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3 years ago

5

Suppose you are playing hockey on a new-age ice surface for which there is no friction between the ice and the hockey puck. You

wind up and hit the puck as hard as you can. After the puck loses contact with your stick, the puck will

1 answer:

BlackZzzverrR [31]3 years ago

4 0

Answer:

<em>Not slow down or speed up</em>.

Explanation:

Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.

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A steam engine absorbs 4 x 105 J and expels 3.5 x 105 J in each cycle. What is its efficiency?

Wewaii [24]

Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J

From the first Law of thermodynamics, obtain useful work performed as
W = Qin - Qout
= 0.5 x 10⁵ J

By definition, the efficiency is
η = W/Qin
= 100*(0.5 x 10⁵/4 x 10⁵)
= 12.5%

Answer: The efficiency is 12.5%

3 0

3 years ago

This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

3 0

2 years ago

A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr

mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

$f' = (\frac{v}{v\pm v_s})f$

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

$v_s$ is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

$v_s = -40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz$

When the train has passed the platform, we have

$v_s = +40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz$

Therefore the overall shift in frequency is

$\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz$

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

$v=\lambda f$

where

v is the speed of the wave

$\lambda$ is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

$\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m$

5 0

3 years ago

The Faraday disk was an early example of what?

Evgen [1.6K]

Electrical generator which operates using a magnetic field. It is the beginning of modern dynamos.

8 0

3 years ago

En la figura los émbolos son de masa despreciable y están en reposo, como se ve en la figura, siendo M=300 kg. Determine: a) la

Leni [432]

The image mentioned is in the attachment

Answer: a) P = 2450 Pa;

b) P = 2940 Pa;

c) F = 4.9 N

Explanation:

a) Pressure is a force applied to a surface of an object or fluid per unit area.

The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:

P = $\frac{F}{A}$

P = $\frac{m.g}{A}$

P = $\frac{300.9.8}{1.2}$

P = 2450 Pa

The pressure generated by the block is P = 2450 Pa.

b) A static liquid can also exert pressure and can be calculated as:

$P_{staticfluid} =$ ρ.g.h

where

ρ is the density of the fluid

h is the depth of the fluid

g is acceleration of gravity

$P_{staticfluid} =$ 600.9.8.0.5

$P_{staticfluid} =$ 2940 Pa

The pressure in the fluid at 50 cm deep is $P_{staticfluid} =$ 2940 Pa.

c) For the system to be in equilibrium both pressures, pressure on the left side and pressure on the right side, have to be the same:

$P_{s} = P_{b}$

$\frac{F}{A_s}$ = $\frac{F_b}{A_b}$

F = $\frac{F_b}{A_b}.A_s$

Adjusting the units, $A_{s}$ = 0.002 m².

F = $\frac{300.9.8.0.002}{1.2}$

F = 4.9 N

The force necessary to be equilibrium is F = 4.9 N.

4 0

3 years ago