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ella [17]
3 years ago
10

In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the

puck and the horizontal board is 0.28, what was the initial speed of thepuck?
Physics
1 answer:
Artist 52 [7]3 years ago
4 0

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s    ................1

we know acceleration will be

a = \frac{-u^2}{2\times S}  

Then we have

Force = mass × acceleration   .................2

m × \frac{-u^2}{2\times S} = -μk mg

The puck decelerates due to the kinetic frictional force μk mg  

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

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25 points! Please someone help me, this seems like a pretty easy question but I can’t seem to get it right, please list the give
elixir [45]

Answer:

1 my brother say that

Explanation:

i know my brother said it

4 0
3 years ago
14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards
quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

8 0
1 year ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
4 years ago
Un objeto se mueve con una rapidez constante de 8 m/s. Esto significa que el objeto: a) Aumenta su rapidez en 8m/s cada segundo
Hatshy [7]

Explanation:

Constant speed means that the object is covering equal distance in equal interval of time. The motion is called uniform motion for such case. In this problem, it is given that an object is moving with a constant speed of 8 m/s. It means that it does not change its speed. It is moving with a single speed constantly.

We can say that it moves 8 meters every second or 800 cm every second.

4 0
4 years ago
Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
Crazy boy [7]

Answer:

Temperature at the exit = 267.3 C

Explanation:

For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

5 0
3 years ago
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