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Virty [35]
3 years ago
6

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a hor

izontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Physics
1 answer:
Leno4ka [110]3 years ago
7 0

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

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Galileo successfully demonstrated that the balls took the same amount of time to reach the ground.

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How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?(gravity is 9.8)
Mekhanik [1.2K]
First, calculate the initial velocity of the dog given with the vertical height and the acceleration due to gravity which is calculated through the equation,
                         2ad = Vo²
Substituting the known values,
                           2(9.8 m/s²)(1.2 m) = V₀²
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The kinetic energy is solved through the equation, 
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Substituting the known values to the latest equation,
                         KE = 0.5 (7.2 kg)(4.85 m/s)²
                          KE = 17.46 J
Thus, the kinetic energy is 17.46 J. 
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Gallium arsenide (GaAs) and gallium phosphide (GaP) are compound semiconductors that have room-temperature band gap energies of
Firdavs [7]
Hence, 0.60+2.25eV is the solution to this problem
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2 years ago
A train is moving along a track with constant speed v1 relative to the ground. A person on the train holds a ball of mass m and
scoundrel [369]

Answer:

(a)0.5m(2v1v2+v2^{2})\\(b)0.5m(v2^{2}-v1^{2})\\(c)0.5m(2v1v2+v2^{2})\\(d)0.5m(v2^{2}-v1^{2})

Explanation:

velocity of ball in train reference = v2

velocity of ball in earth reference = v1+v2

(a)

Kinetic energy is given by 0.5mv^{2} where m and v are the mass and velocity of object respectively.

Change in kinetic energy is given by subtracting initial kinetic energy from the final kinetic energy. In this case

Initial kinetic energy= 0.5mv1^{2}

Final kinetic energy= 0.5m (v1+v2)^{2}=0.5m(v1^{2}+2v1v2+v2^{2})

Change in kinetic energy=0.5m (v1+v2)^{2}-0.5mv1^{2}=0.5m((v1+v2)^{2})-v1^{2}=0.5m(2v1v2+v2^{2}

(b)

Change in velocity in train reference will be

Initial kinetic energy= 0.5mv1^{2}

Final kinetic energy= 0.5mv2^{2}

Change in kinetic energy=0.5m(v2^{2}-v1^{2})

(c)

Work done, W = change in kinetic energy=0.5m (v1+v2)^{2}-0.5mv1^{2}=0.5m((v1+v2)^{2})-v1^{2}=0.5m(2v1v2+v2^{2}

(d)

Work done, W = change in kinetic energy=0.5m(v2^{2}-v1^{2})

4 0
3 years ago
Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic
Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

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Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by \frac{1}{2}kx^{2} where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

                                                              ⇒\frac{1}{2}k(\frac{w}{k})^{2}

                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

4 0
3 years ago
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