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2 years ago

3. Thekinetic energy of an object of

Physics

1 answer:

lawyer [7]2 years ago

7 0

Answer:

1. 100 J

2. 225 J

Explanation:

We'll begin by calculating the mass of the object. This can be obtained as follow:

Velocity (v) = 5 ms¯¹

Kinetic energy (KE) = 25 J

Mass (m) =?

KE = ½mv²

25 = ½ × m × 5²

25 = ½ × m × 25

25 = 25m / 2

Cross multiply

25m = 25 × 2

25m = 50

Divide both side by 25

m = 50 / 25

m = 2 Kg

1. Determination of the kinetic energy when the velocity is doubled.

Mass (m) = 2 Kg

Velocity (v) = double the initial velocity

= 2 × 5 ms¯¹

= 10 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 10²

KE = ½ × 2 × 100

KE = 100 J

2. Determination of the kinetic energy when the velocity increased three times.

Mass (m) = 2 Kg

Velocity (v) = three times the initial velocity

= 3 × 5 ms¯¹

= 15 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 15²

KE = ½ × 2 × 225

KE = 225 J

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dem82 [27]

Answer:

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2 years ago

A yellow and green car traveled 400 miles to Dayton, OH. The green car made the trip in 10 hours. The yellow arrived in 8 hours.

elixir [45]

Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

Explanation:

The speed of each car is defined as:

$v=\frac{d}{t}$

where d is the distance traveled by the car and t is the time taken.

For the yellow car, d=400 mi and t=8 h, so its speed is

$v=\frac{400 mi}{8 h}=50 mph$

For the green car, d=400 mi and t=10 h, so its speed is

$v=\frac{400 mi}{10 h}=40 mph$

So, the correct choice is

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

4 0

3 years ago

Read 2 more answers

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :