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3 years ago

A car skids to a stop. What happens to its kinetic energy?

Physics

1 answer:

Orlov [11]3 years ago

4 0

Answer:

a) it turns into thermal energy

Explanation:

Due to the law of energy conservation, an object cannot just "lose" its kinetic energy, it must become another type of energy.

This eliminates option b. because energy of any type cannot disappear.

About option c, we have that the kinetic energy becomes potential energy, but the potential energy of an object is due to its height above the ground, so we cannot have potential energy if the tires are on the floor, we discard option c.

Also, kinetic energy is the energy an object has due to its motion, this means that if the object is not moving, it cannot have kinetic energy.

This eliminates option d.

The correct option is a) it turns into thermal energy.

The option meets the characteristic that the kinetic energy does not disappear, it is transformed into another type of energy. In this case it is thermal energy or heat, generated by friction between the tires and the floor.

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A 1,383 kg purple car is driving southbound on a road and collides with a 1,827 kg orange car, that was traveling 31.87 m/s east

user100 [1]

Answer:

Explanation:

We shall apply work energy theorem to calculate the initial velocity just after the collision .

Their kinetic energy will be equal to work done by friction .

force of friction = μ mg , where μ is coefficient of friction , m is total mass and g is acceleration due to gravity

force = .463 x 3210 x 9.8

= 14565.05 N

work done = force x displacement

= 14565.05 x 14.54 = 211775.88 J

now applying work energy theorem

1/2 m v² = 211775.88 , m is composite mass , v is velocity just after the collision

.5 x 3210 x v² = 211775.88

v² = 131.94

v 11.48 m /s

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3 years ago

A frame of reference is a point of set or points that.....

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Describes the displace, velo, and accel of an object

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3 years ago

The perceived pitch of a tone is largely determined by its

soldier1979 [14.2K]

The perceived pitch of a tone is largely determined by its frequency

<h3>What is frequency ?</h3>

The frequency of a repeated event is its number of instances per unit of time. In some cases, it is also referred to as temporal frequency or ordinary frequency to underline differences with spatial and angular frequencies, respectively. One (event) per second is equal to one hertz (Hz), which is how frequency is stated.

The period is the reciprocal of the frequency since it is the length of time for one cycle in a repeating occurrence. If a heart beats 120 times per minute (2 hertz), for instance, its period, T—the space between beats—is half a second. In science and engineering, frequency is a crucial metric for describing the temporal rate of change seen in oscillatory systems.

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2 years ago

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has dr

Wittaler [7]

Answer:

A. ) K =126. 7 J

B. ) h= 91.1 m.

Explanation:

Assuming no air resistance, once released by the pitcher, the speed must keep constant through all the trajectory, so the kinetic energy of the ball can be expressed as follows:

$K = \frac{1}{2}*m*v^{2} = \frac{1}{2}*0.142 kg*(42.24m/s)^{2} = 126.7 J (1)$

Neglecting air resistance, total mechanical energy must be the same at any point, so, if we choose the ground level as the zero reference level for the gravitational potential energy, and assuming that the ball attains this kinetic energy just before striking ground, this value must be equal to the gravitational potential energy just before be dropped, so we can write the following equality:

$U_{o} = K_{f} = 126. 7 J (2)$

⇒ m*g*h = 126. 7 J

Solving for h, we get:

$h = \frac{K_{f}}{m*g} = \frac{126.7J}{0.1420kg*9.8m/s2} = 91.1 m (3)$

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3 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

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3 years ago