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I am Lyosha [343]
3 years ago
15

A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk a

nd the concrete court is 0.31, how far does the disk go before it comes to a stop?
Physics
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

5.53 m

Explanation:

v_{o} = initial speed of  shuffleboard disk = 5.8 ms^{-1}

v_{f} = final speed of  shuffleboard disk = 0 ms^{-1}

\mu = Coefficient of kinetic friction between the disk and concrete court = 0.31

acceleration due to friction is given as

a = - \mu g\\a = - (0.31) (9.8)\\a = - 3.04 ms^{-2}

d = distance traveled by disk before it stops

using the kinematics equation that fits the above the data, we have

v_{f}^{2} = v_{o}^{2} + 2 a d\\0^{2} = 5.8^{2} + 2 (- 3.04) d\\d = 5.53 m

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Scenario
zepelin [54]

Answer:

 t = 23.255 s,   x = 2298.98 m,    v_y = - 227.90 m / s

Explanation:

After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

         v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s

         y₀ = 2650 m

Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement

Y axis  

       y = y₀ + v₀ t - ½ g t²

the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero

       0 = y₀ + 0 - ½ g t²

       t = \sqrt{  \frac{2 y_o}{g} }

       t = √(2 2650/ 9.8)

       t = 23.255 s

Therefore, for the cargo to reach the desired point, it must be launched from a distance of

       x = v₀ₓ t

       x = 111.76 23.255

       x = 2298.98 m

at the point and arrival the speed is

        vₓ = v₀ₓ = 111.76

     

vertical speed is

         v_y = v_{oy} - gt

          v_y = 0 - gt

          v_y = - 9.8 23.25 555

         v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement

7 0
2 years ago
Why are the Soviets ahead in the race to capture the V-2 rocket and Werner von Braun?
Dahasolnce [82]

The second world war, and its war weapons, such as the v-2 rockets, had a great impact on the world until today, to answer this question we need that...

<h3>V-2 factory </h3>

On April 11, 1945, US troops took the town of Bleicherode, in the Kohnstein region, where the V-2 factory was located. From there about 100 complete V-2s and thousands of parts and equipment were "captured" as war loot and transferred to the United States, where they formed the basis for practical studies of the missile defense program.

With this information, we can say that because it was a base discovered by US troops, none of the alternatives is correct, as it was not the Soviets who discovered it, and that the base was also located in the central part of Germany.

<u>The </u><u>v-2 factory</u><u> was located in </u><u>Kohnstein, central germany,</u><u> by this claim, </u><u>none of the alternatives is correct.</u>

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5 0
2 years ago
Help me, 100 points to answer right, answer without context will be reported
dlinn [17]
734 is the answer for the chronic blood exchange service of new france
7 0
2 years ago
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

\Sigma F = F_{E}-W = 0 (Eq. 1)

Where:

F_{E} - Electrostatic force exerted on human, measured in Newton.

W - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

q\cdot E-m\cdot g = 0

q = \frac{m\cdot g}{E} (Eq. 2)

E - Electric field, measured in Newtons per Coloumb.

m - Mass, measured in kilograms.

g - Gravity acceleration, measured in meters per square second.

q - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that m = 60\,kg, g = 9.807\,\frac{m}{s^{2}} and E = -150\,\frac{N}{C}, the charge that a 60-kg human must have to overcome weight is:

q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }

q = -3.923\,C

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

F = \kappa \cdot \frac{q^{2}}{r^{2}} (Eq. 3)

Where:

\kappa - Electrostatic constant, measured in Newton-square meter per square Coulomb.

q - Electric charge, measured in Coulomb.

r - Distance between two people, measured in meters.

If we know that \kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q = -3.923\,C and r = 100\,m, then the force of repulsion between two people is:

F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]

F = 13.851\times 10^{6}\,N

The force of repulsion between two people is 13.851\times 10^{6} newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0
3 years ago
When a cyclist moves downhill without pedalling, what type of energy does he gain?
Vesna [10]

Answer:

He is gaining kinetic energy and losing potential energy

Explanation:

3 0
2 years ago
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