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natali 33 [55]
3 years ago
10

A 0.060-kg ice hockey puck comes toward a player with a high speed. The player hits it directly back softly with an average forc

e of 1.50 x 10^3 N. The hockey stick is in contact with the ball for 1.20 ms, and the ball leaves the stick with a velocity of 8.00 m/s. Let the direction of the force be the + x direction. Find the following (note: be careful with the sign/direction of the values):_______. 1. The final momentum of the ball 2. The impulse on the ball 3. The initial velocity of the ball
Physics
1 answer:
kramer3 years ago
8 0

Answer:u=-22 m/s

Explanation:

Given

mass of puck m=0.06\ kg

Average force f_{avg}=1.5\times 10^3\ N

time of contact t=1.2ms=1.2\times 10^{-3}\ s

puck leaves with a velocity of v=8\ m/s

We know impulse is F_{avg}\Delta t=\text{change in momentum}

therefore

1.5\times 10^3\times (1.2\times 10^{-3})=P_f-P_i

P_i=0.06\times 8-1.8

P_i=0.48-1.8=-1.32\ kg-m/s

Final momentum P_f=m\times v_f

P_f=0.06\times 8

P_f=0.48\ kg-m/s

Impulse on the ball =F_{avg}\Delta t

Impulse=1.5\times 10^3\times 1.2\times 10^{-3}=1.8\ N-s

Initial velocity is given by

u=\frac{P_i}{m}=\frac{-1.32}{0.06}

u=-22\ m/s

i.e. initially ball is moving towards -x-axis

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A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
2 years ago
A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha
kifflom [539]

Answer:

x_{distance}=10.07km

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

V_{avg}=(x_{distance} )/(t_{time} )\\ x_{distance}=V_{avg}*(t_{time} )\\x_{distance}=(19.0km/h)*(0.530h)\\x_{distance}=10.07km

3 0
3 years ago
An acorn with a mass of 0.300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.oo m/s?
vladimir1956 [14]

Answer: 3.75 joules

Explanation:

Given that:

Mass of acorn = 0.300 kilograms

velocity = 5.oo m/s

Kinetic energy = ?

Since, kinetic energy is the energy possessed by a moving object, its value depends on the mass M and velocity V of the acorn.

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 0.300kg x (5.00m/s)^2

= 0.5 x 0.3kg x (5.00m/s)^2

= 0.15 x (5.00m/s)^2

= 3.75 joules

Thus, the kinetic energy of the falling acorn is 3.75 joules

5 0
2 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
The loudness of a sound is the waves amplitude <br><br><br><br> True or false
kakasveta [241]

Answer:

true i think

Explanation:

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound, and a smaller amplitude means a softer sound. In Figure 10.2 sound C is louder than sound B. The vibration of a source sets the amplitude of a wave.

4 0
2 years ago
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