All categories

3 years ago

What is the acceleration of a car that travels in a straight line?

2 answers:

6 0

Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.

hope this helps!! ✨

Bond [772]3 years ago

4 0

Average and instantaneous (by contrast there is also centripetal for angular motion)

You might be interested in

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori

EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

$x - x_o = u_xt$

$\mathtt{x = u_xt \ \ \ since (x_o = 0)}$ ---- (1)

the equation of the motion y is :

$\mathtt{y - y_o =u_yt - 0.5 gt^2}$

$\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}$

$\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }$

$\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}$

$\mathtt{1 = 5.14t - 4.9t^2}$

$\mathtt{4.9t^2 - 5.14t +1 = 0}$

By using the quadratic formula, we have;

$\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }$

where;

a = 4.9, b = -5.14 c = 1

$= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }$

$= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }$

$= \mathbf{ 0.791 \ \ OR \ \ 0.258} }$

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

$\mathtt{x = u_x(0.258)}$

$\mathtt{x = ucos 40^0 (0.258)}$

$\mathtt{x = 8 \ cos 40^0 (0.258)}$

$\mathbf{x = 1.581 \ m}$

Thus, the child is 1.581 m far from the fence

6 0

2 years ago

State the Newton 2nd law of motion and also prove that F= ma

saw5 [17]

Explanation:

F = ma is the formula of Newton's Second Law of Motion. Newton's Second Law of Motion is defined as Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.

...

3 0

3 years ago

Which of the following electrical components is a temporary electrical energy storage device?

7nadin3 [17]

Answer:

A capacitor

Explanation:

Because it can store electric energy when disconnected from its charging circuit. Commonly used in electronic devices to maintain power supply while batteries change.

Hope this helps! :)

6 0

2 years ago

E= (500.0lm) 4 (100)

drek231 [11]

Answer:

didn't understand your question

7 0

3 years ago

A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t

koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

$\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N - m\cdot g = 0$ (2)

Where:

$N$ - Normal force from the ground on the object, measured in newtons.

$m$ - Mass of the object, measured in newtons.

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

$v$ - Linear speed of rotation of the disk, measured in meters per second.

$R$ - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{g\cdot R}$

If we know that $v = 0.8\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 0.75\,m$ , then the coefficient of static friction between the object and the disk is:

$\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}$

$\mu_{s} = 0.087$

The coefficient of static friction between the object and the disk is 0.087.

5 0

2 years ago