Ask question

Ask question

All categories

3 years ago

10

Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c

ar and apply forces to it. Use the format of "if . . . then . . . because . . .” and be sure to answer the lesson question "How can Newton's laws be experimentally verified?” specific to Newton’s second law.

2 answers:

skelet666 [1.2K]3 years ago

7 0

Sample Response: If force is applied to a car, then its acceleration will change proportionally, as predicted by Newton’s second law, F = ma.

Temka [501]3 years ago

6 0

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

You might be interested in

How did particles in the solar nebula eventually form earth

Sladkaya [172]

Answer:

At the high temperatures of the inner solar nebula, the small proto-planets were too hot to hold the volatile gases that dominated the solar nebula. These proto-planets were Earth, Mars, Venus, and Mercury.

Explanation:

The materials that accreted into the early Earth were probably added piecemeal, without and particular order. The early earth was very hot from gravitational compression, impacts and radioactive decay; the earth was partially molted. The denser metallic liquids sank to the center of the Earth and less denser silicate liquids rose to the top. In this way the Earth differentiated very quickly into a metallic, mostly iron core and a rocky silicate mantle.

5 0

3 years ago

Clara rushes 30m to a truck then turns and walks back. Total travel is 120s what is her average velocity?

navik [9.2K]

Taking into account the definition of velocity, Clara's average velocity is 0.5 m/s.

<h3>Definition of velocity</h3>

Velocity is a physical magnitude that relates the displacement of an object, the time it takes to make this change in position and direction. So it is considered a vector magnitude.

In other words, the velocity can be defined as the amount of space traveled per unit of time with which a body moves, considering the direction, and can be calculated using the expression:

velocity= distance traveled÷ time

<h3>Average velocity of Clara</h3>

Clara rushes 30 m to a truck then turns and walks back. Total travel is 120s. Then, you know:

distance traveled= 30m rushing + 30m walking back= 60 m
time= 120 s

Replacing in the definition of velocity:

velocity= 60 m÷ 120 s

Solving:

<u><em>velocity= 0.5 m/s</em></u>

Finally, Clara's average velocity is 0.5 m/s.

Learn more about velocity:

brainly.com/question/4418301

brainly.com/question/24990123

brainly.com/question/14244558

brainly.com/question/25288184

#SPJ1

5 0

2 years ago

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 $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

3 years ago

Acceleration is defined as the rate of change for which of the following? Question 4 options: time position velocity displacemen

vovangra [49]

Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.

<h3>What is Acceleration?</h3>

It is defined as the rate of change in velocity.

It can also be defined as the rate of change in position in a particular direction.

$a = \dfrac {v-u}t\ \ \ \ \rmor}\\\\a = \dfrac {\Delta v }t$

Where,

$a$ - acceleration

$\Delta v$ - change in velocity

$t$ - time

Therefore, acceleration can be defined as the rate of change in the velocity of an object.

Learn more about Velocity:

brainly.com/question/2239252

4 0

2 years ago