(Please Help! Match the sentence to the number down below! Please I'll give brainlist and points!)

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

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

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

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

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

3 years ago

In science, does acceleration happen when an object is slowing down?

Irina18 [472]

Answer:

According to our principle, when an object is slowing down, the acceleration is in the opposite direction as the velocity. Thus, this object has a negative acceleration. In Example D, the object is moving in the negative direction (i.e., has a negative velocity) and is speeding up.

7 0

2 years ago

The illustration in figure below shows a uniform metre rule weighing 30 N pivoted on a wedge placed under the 40 cm mark and car

Nitella [24]

Answer:

W = 30 N

Explanation:

Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.

$(70\ N)(40\ cm - 10\ cm)-(30\ N)(50\ cm-40\ cm)-(W)(100\ cm - 40\ cm) = 0\\W(60\ cm) = (70\ N)(30\ cm)-(30\ N)(10\ cm)\\\\W = \frac{1800\ N.cm}{60\ cm}$

W = 30 N

6 0

2 years ago

18°C = _____ -255 K 0 K 18 K 291 K

Nadya [2.5K]

18°C answer is 291 k

hope it help

4 0

3 years ago

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Which of the following is true for gravitational force?

algol13

The answer is b increases in increase of mass because the heavier the object is the more gravitional pull it will have. For example, what will fall faster? A 1 pound object or a 10 pound obect.

8 0

3 years ago

Read 2 more answers