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slamgirl [31]
2 years ago
7

A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp

eed of 6 m/s. If the rink is 61 meters long, how fast is the puck moving when it hits the far wall?
Physics
1 answer:
Dimas [21]2 years ago
5 0

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

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Carbon-14 is used to determine the time an organism was living. The amount of carbon-14 an organism has is constant with the atm
lutik1710 [3]

Answer:

The age of the organism is approximately 11460 years.

Explanation:

The amount of carbon-14 decays exponentially in time and is defined by the following equation:

\frac{n(t)}{n_{o}} = e^{-\frac{t}{\tau} } (1)

Where:

n_{o} - Initial amount of carbon-14.

n(t) - Current amount of carbon-14.

t - Time, measured in years.

\tau - Time constant, measured in years.

Then, we clear the time within the formula:

t = -\tau \cdot \ln \frac{n(t)}{n_{o}} (2)

In addition, time constant can be calculated by means of half-life of carbon-14 (t_{1/2}), measured in years:

\tau = \frac{t_{1/2}}{\ln 2}

If we know that \frac{n(t)}{n_{o}} = 0.25 and t_{1/2} = 5730\,yr, then the age of the organism is:

\tau = \frac{5730\,yr}{\ln 2}

\tau \approx 8266.643\,yr

t = -(8266.643\,yr)\cdot \ln 0.25

t \approx 11460.001\,yr

The age of the organism is approximately 11460 years.

8 0
3 years ago
Read 2 more answers
Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?
snow_tiger [21]
Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2                             ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2




4 0
3 years ago
Read 2 more answers
the object is in _________ when the vector sum of the force is acting on the object is equal to zero​
Norma-Jean [14]

Answer:

Equilibrium

Explanation:

An object is in equilibrium when the vector sum of the force acting on the object is equal to zero.

A body in equilibrium is at state of rest of rest or in motion with no external force acting on it.

  • The resultant of all forces acting on the body is zero.
  • In this case there is no net force and the body will be at rest.
7 0
3 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0
3 years ago
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lukranit [14]

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

8 0
3 years ago
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