Ask question

Ask question

All categories

3 years ago

15

A truck accelerates From being stopped to 18 m/s in 5.5 seconds. A bus accelerates from being stopped to 24 m/s in 6.0 seconds.

Which vehicle has a higher acceleration

1 answer:

Mars2501 [29]3 years ago

5 0

Answer:

The bus

Explanation:

a = (v-u)/t

Where

a = acceleration

v = final velocity

u = initial velocity

t = time taken

For truck to get its acceleration,

a = (18-0)/5.5 = 3.27 ms⁻²

For bus to get its acceleration,

a = (24-0)/6 = 4 ms⁻²

As 4 > 3.27 bus has a greater acceleration.

You might be interested in

Two trumpet players are trying to tune their instruments. When they are in tune, they will both be playing the same note and no

marusya05 [52]

Answer:

The second trumpeter will be playing at frequency = 515 Hz

Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s.

Beat frequency = 20/4 = 5 Hz

Beat frequency = F2 - F1

5 = 520 - F1

F1 = 520 - 5

F1 = 515 Hz

Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency

8 0

3 years ago

How to represent milligram in kilogram by standard formula?

Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take $\frac{1}{1,000,000}$ which equals 0.000001 kg

4 0

3 years ago

A<br> B <br> C <br> D<br><br> Plz help me.

Montano1993 [528]

Answer:

The correct option is;

${}$ Man doing most work ${}$ Box gaining the most energy

D ${}$ Y Q

Explanation:

The given parameters of the question are;

The distance the box P is pushed by the Man X = 0

The force the Man X applies to the box P = x N

The distance the box Q is lifted by the Man Y = h > 0 meters

The minimum force the Man Y applies to the box Q = W, the weight of the box

Work done = Force × Distance

Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height = W × h) + 1/2 × Mass × Velocity²

The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)

Therefore, Kinetic energy = 0 J

The work done by Man X = 0 × x = 0 J

The energy gained by the box P = W × 0 = 0 J

The work done by Man Y = W × h = W·h J

The energy gained by the box P = W × h = W·h J

We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J

The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J

Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.

8 0

3 years ago

4) A satellite, mass m, is in circular orbit (radius r) around the earth, which has mass ME and radius Re. The value of r is lar

defon

<h2>Answers:</h2>

(a) The kinetic energy of a body is that energy it possesses due to its movement and is defined as:

$K=\frac{1}{2}m{V}{2}$ (1)

Where $m$ is the mass of the body and $V$ its velocity.

In this specific case of the satellite, its kinetic energy $K_m$ taking into account its mass $m$ is:

$K_{m}=\frac{1}{2}m{V}^{2}$ (2)

On the other hand, the velocity of a satellite describing a circular orbit is constant and defined by the following expression:

$V=\sqrt{G\frac{ME}{r}}$ (3)

Where $G$ is the gravity constant, $ME$ the mass of the Earth and $r$ the radius of the orbit <u>(measured from the center of the Earth to the satellite). </u>

Now, if we substitute the value of $V$ from equation (3) on equation (2), we will have the final expression of the kinetic energy of this satellite:

$K_{m}=\frac{1}{2}m{\sqrt{G\frac{ME}{r}}}^{2}$ (4)

Finally:

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5) >>>>This is the kinetic energy of the satellite

(b) According to Kepler’s 2nd Law applied in the case of a circular orbit, its Period $T$ is defined as:

$T=2\pi\sqrt{\frac{r^{3}}{\mu}}$ (6)

Where $\mu$ is a constant and is equal to $GME$ . So, this equation in these terms is written as:

$T=2\pi\sqrt{\frac{r^{3}}{GME}}$ (7)

As we can see, <u>the Period of the orbit does not depend on the mass of the satellite </u> $m$ , it depends on the mass of the greater body (the Earth in this case) $ME$ , the radius of the orbit and the gravity constant.

(c) The gravitational force described by the law of gravity is a central force and therefore is <u>a conservative force</u>. This means:

1. The work performed by a gravitational force to move a body from a position A to a position B <u>only depends on these positions and not on the path followed to get from A to B. </u>

2. When the path that the body follows between A and B is a c<u>losed path or cycle</u> (as this case with a <u>circular orbit</u>), <u>the gravitational work is null or zero</u>.

<h2>This is because the gravity force that maintains an object in circular motion is a centripetal force, that is, <u>it always acts perpendicular to the movement</u>. </h2>

Then, in the case of the satellite orbiting the Earth in a circular orbit, its movement will always be perpendicular to the gravity force that attracts it to the planet, at each point of its path.

(d) The total Mechanical Energy $E$ of a body is the sum of its Kinetic Energy $K$ and its Potential Energy $P$ :

$E=K+P$ (8)

But in this specific case of the circular orbit, its kinetic energy will be expresses as calculated in the first answer (equation 5):

$K_{m}=\frac{1}{2}Gm\frac{ME}{r}$ (5)

And its potential energy due to the Earth gravitational field as:

$P_{m}=-G\frac{mME}{r}$ (9)

This energy is negative by definition.

So, the total mechanical energy of the orbit, also called the Orbital Energy is:

$E=\frac{1}{2}Gm\frac{ME}{r}+(- G\frac{mME}{r})$ (10)

Solving equation (10) we finally have the Orbital Energy:

$E=-\frac{1}{2}mME\frac{G}{r}$ (11)

At this point, it is necessary to clarify that a satellite (or any other celestial body) orbiting another massive body, can describe one of these types of orbits depending on its Orbital Total Mechanical Energy $E$ :

-When $E=0$ :

We are talking about an <u>open orbit</u> in which the satellite escapes from the attraction of the planet's gravitational field. The shape of its trajectory is a parabola, fulfilling the following condition:

$K_{m}=-P_{m}$

Such is the case of some comets in the solar system.

-When $E>0$ :

We are also talking about <u>open orbits</u>, which are hyperbolic, being $K_{m}>P_{m}$

<h2>-When $E$ : >>>><u>This case</u></h2>

We are talking about <u>closed orbits</u>, that is, the satellite will always be "linked" to the gravitational field of the planet and will describe an orbit that periodically repeats with a shape determined by the relationship between its kinetic and potential energy, as follows:

-Elliptical orbit: Although $E$ is constant, $K_m$ and $P_m$ are changing along the trajectory .

-Circular orbit: When at all times both the kinetic energy $K_m$ and the potential $P_m$ remain constant, resulting in a total mechanical energy $E$ as the one obtained in this exercise. This means that the speed is constant too and <u>is the explanation of why this Energy has a negative sign. </u>

3 0

4 years ago

use the instantaneous center of zero velocity to determine the angular velocity !ab of link ab and the velocity vb of collar b f

allsm [11]

Instantaneous center:

It is the center about a body moves in planer motion. The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body. About this center every particle of a body rotates.

From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.

To know more about velocity, refer: brainly.com/question/12413963

#SPJ4

7 0

1 year ago