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2 years ago

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A 75.3 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 8.1 m starting

from rest, its speed is 7.1 m/s. Find the magnitude of the net force on the bobsled. Answer in units of N

1 answer:

dusya [7]2 years ago

8 0

Answer:

233.43 N

Explanation:

Force: This is the product of mass and acceleration of a body.

The formula for force is given as,

F = ma .............. Equation 1

Where F = Net force on the bobsled, m = mass of the bobsled, a = acceleration of the bobsled

We can look for a using the equation of motion

v² = u² + 2as.............. Equation 2

Where V = final velocity, u = initial velocity, a = acceleration, s = distance.

making a the subject of the equation,

a = (v²-u²)/2s................... Equation 3

Given: v = 7.1 m/s, u = 0 m/s ( from rest), s = 8.1 m.

Substitute into equation 3

a = (7.1²-0²)/(2×8.1)

a = 50.41/16.2

a = 3.1 m/s²

Also given: m = 75.3 kg

Substitute into equation 1

F = 75.3×3.1

F = 233.43 N

Hence the net force on the bobsled = 233.43 N

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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Explanation:

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2 years ago

Describe a ball's motion as it rolls up a slanted

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The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

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When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

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Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

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$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

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With the value of the tension and the velocity you can find the mass density:

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2 years ago

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bija089 [108]

Answer:

Nodes.

Explanation:

Nodes are a point that are on a standing wave that never move.

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2 years ago

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