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Alex777 [14]
3 years ago
10

A car accelerates from a resting position at a constant 5 meters per second per second. What does the graph of acceleration vers

us time look like

Physics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

You just told us that the acceleration is constant, and we believed you. So if we draw a graph of it, and it never gets any higher or lower than 5 m/s², then the graph is a straight horizontal line.

kvv77 [185]3 years ago
4 0

Explanation:

It is given that, a car accelerates from a resting position at a constant 5 meters per second per second. When the car moves with constant velocity, then the change in velocity is equal to 0.

The attached figure shows the acceleration time graph when the car accelerates with constant velocity. The line is parallel to the time axis or x axis.

So, the acceleration versus time graph look like line parallel to x axis. Hence, this is the required solution.

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I. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maxi
nalin [4]

Answer:

Negative

Explanation:

If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.

Hope this helped. :)

6 0
3 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
What is the proper battery cable connection when jumping two automotive batteries? (a) negative to negative / positive to positi
Vladimir [108]
<span>The proper </span><span>battery cable connection when jumping two automotive batteries is :  </span><span>(a) negative to negative / positive to positive. 

</span><span>Connect the red (positive) cable from the car with the bad battery to the red (positive) on the good battery. </span>

<span>Then connect the black (negative) from the good battery to a grounding point on the other car which should be tightened and metal should be clean.
</span>
<span>Once the car with bad battery has started, the removal of the cable should be in the opposite order. The Red (positive) which was the the First Cable to go on should be the last cable to be taken off.</span>


3 0
3 years ago
Engineers are designing a curved section of a highway. If the radius of curvature of the curve is 194 m, at what angle should th
Brums [2.3K]

Answer:

The banking angle is 23.84 degrees.

Explanation:

Given that,

Radius of the curve, r = 194 m

Speed of the car, v = 29 m/s

On the banked curve, the centripetal force is balanced by the force of friction such that,

mg\ tan\theta=\dfrac{mv^2}{r}

tan\theta=\dfrac{v^2}{rg}

tan\theta=\dfrac{(29)^2}{194\times 9.8}

\theta=23.84^{\circ}

So, the banking angle is 23.84 degrees. Hence, this is the required solution.    

5 0
3 years ago
Magnetism/ magnetic field ana magnetic forces
katovenus [111]

Answer:

Magnetism is a physical phenomenon that manifests itself in a force acting between magnets or other magnetized or magnetisable objects, and a force acting on moving electric charges, such as in current-carrying cables. The force action takes place by means of a magnetic field, which is generated by the objects themselves or otherwise. There are natural and artificial magnets. All magnets have two poles called the north pole and the south pole. The north pole of one magnet repels the north pole of another magnet and attracts the south pole of another magnet; the same with south poles.  

6 0
3 years ago
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