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3 years ago

is the following sentence true or false? the faster the particles of a substance are moving, the more energy they have.

Physics

2 answers:

svetlana [45]3 years ago

8 0

I think true. I'm pretty sure, but check w/ others too.

rosijanka [135]3 years ago

3 0

The anwer is definitely true. That is what I think.

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A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

The maximum height a typical person can jump is about 60cm (0.6m). By how much does the gravitational potential energy increase

sasho [114]

The gravitational potential energy will increase by 423.36 J

<h3>How to determine the potential energy at ground level</h3>

Mass (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 0 m
Potential energy at ground level (PE₁) =?

PE = mgh

PE₁ = 72 × 9.8 × 0

PE₁ = 0 J

<h3>How to determine the potential energy at 60 cm (0.6 m)</h3>

Mass (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 0.6 m
Potential energy at 60 cm (0.6 m) (PE₂) =?

PE = mgh

PE₂ = 72 × 9.8 × 0.6

PE₂= 423.36 J

<h3>How to determine the change in potential energy </h3>

Potential energy at ground level (PE₁) = 0 J
Potential energy at 60 cm (0.6 m) (PE₂) = 423.36 J
Change in potential energy =?

Change in potential energy = PE₂ - PE₁

Change in potential energy = 423.36 - 0

Change in potential energy = 423.36 J

Learn more about energy:

brainly.com/question/10703928

#SPJ1

8 0

2 years ago

Calculate the speed of a train that traveled 810 kilometers in 9 hours.

Naya [18.7K]

Answer:

90 kilometers an hour

let me know if I'm wrong

8 0

3 years ago

Definition of distance travelled?

Mekhanik [1.2K]

Answer:

distance traveled is a total length of the path traveled between two positions.

6 0

3 years ago

Read 2 more answers

1. The picture below shows Jamal pushing with a 100-Newton (N) force on a large box. Neither

r-ruslan [8.4K]

Answer:

100 newtons

Explanation:

Given,

Jamal pushing a large box by a force, F = 100 N

Work done on the large box is, W = 0

It is because the applied force is less than the force of the friction between the two surfaces.

Yet, there will be a force that is exerted by the large box on Jamal.

According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

$F_{a} =-F_{r}$

Hence, If the action force is 100 N, then the reaction force should be in 100 N

4 0

4 years ago

Read 2 more answers