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3 years ago

10

The spring of a toy car stores 10 J of potential energy. Only 8 J of energy changes to kinetic energy as the car moves. What hap

pens to the remaining 2 J of energy?

1 answer:

son4ous [18]3 years ago

3 0

Answer:

It has been converted into thermal energy due to friction

Explanation:

According to the law of conservation of energy, energy cannot be created nor destroyed, but only transformed from one form into another.

Applied to this problem, it means that the total initial energy of the spring-toy system must be conserved.

Therefore:

- At the beginning, the total energy stored in the spring is 10 J

- After the toy is released, the total energy must still be 10 J.

In reality, we are told that the kinetic energy of the car is only 8 J. The other 2 J have not been destroyed, but they have been converted into thermal energy, due to the presence of frictional forces that act against the motion of the toy car.

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Two objects of same material are travelling near you. Object A is a 1.9 kg mass traveling 8 m/s; object B is a 2 kg mass traveli

Alex73 [517]

To determine the object which could give the greatest impact we will apply the concept of momentum. The object that has the highest momentum will be the object that will impact the strongest. Our values are

Mass of Object A

$m_A=1.9 kg$

Velocity of object A

$v_A=8ms$

Mass of object B

$m_B=2 kg$

Velocity of object B

$v_B=5ms$

The general formula for momentum is the product between mass and velocity, then

$p = mv$

For each object we have then,

$p_A=m_Av_A=1.9 kg(8ms)=15.2kg \cdot m/s$

$p_B=m_Bv_B=2 kg(5ms)=10kg \cdot m/s$

Since the momentum of object A is greater than that of object B, then object A will make you feel force upon impact.

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3 years ago

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.

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3 years ago

Someone help please by providing work and answers please :)

cupoosta [38]

There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.

The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!

it's a shortcut.

Let's split up the area into a triangle and rectangle:

Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m

Total distance = 40 m!

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<span>160 Joules

For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance: 80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.

So the cart has a final kinetic energy of 160 Joules.</span>

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2 years ago

A 0.240 kg potato is tied to a string with length 1.90 m, and the other end of the string is tied to a rigid support. The potato

Luba_88 [7]

Explanation:

Below is an attachment containing the solution.

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2 years ago

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