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3 years ago

Describe how the velocity and acceleration of a skydiver changes as she falls from the plane back to the ground

2 answers:

8 0

The velocity and acceleration of the skydiver will increase as she falls to the ground. According to Newton’s Law of gravity.

lidiya [134]3 years ago

8 0

Answer:

Explanation:

The velocity and acceleration of a skydiver increases as she falls from the plane back to the ground. When she will come more closer to earth the force of gravity will increase on her and her velocity will increase so the acceleration. According to the newtons law of gravitation gravity will increase as the distance between the two body decreases.

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While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile's acceleration? (Reme

slava [35]

Answer:

Acceleration = -0.75 m/s²

Explanation:

Given the following data;

Initial velocity = 24 m/s

Final velocity = 15 m/s

Time = 12 seconds

To find the acceleration of the automobile;

Acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac {final \; velocity - initial \; velocity}{time}$

Substituting into the formula, we have;

$Acceleration = \frac{15 - 24}{12}$

$Acceleration = \frac{-9}{12}$

Acceleration = -0.75 m/s²

Therefore, the automobile is decelerating because its final velocity is lesser than its initial velocity, leading to a negative value.

5 0

3 years ago

How does the sun's gravity and the earth inertia keep us orbiting in the solar system

scZoUnD [109]

<span>Inertia keeps us orbiting because any object with mass has the tendency to resist changes to their direction and speed of movement. Combine that with the interaction of the gravitational attraction of the sun, and that is what keeps Earth in orbit. The sunâ€™s gravitational force is one that is proportional to Earthâ€™s mass, and it acts in a way that is almost exactly perpendicular to Earthâ€™s motion. This keeps Earth from spinning into the sun or far away from it.</span>

6 0

4 years ago

11 e) Can a conductor be given limitless charge? Obtain the equivalent resistance of several resistors if (a) they are in series

horsena [70]

Answer:

(e) no

(a) Rs = R' + R'' + R'''

(b) 1/Rp = 1/R' + 1/R'' + 1/R'''

Explanation:

11 e)

Practically it is not possible to give limitless charge to a conductor. It depends to the number of valence electrons.

(a) When the three resistances R'. R'' and R''' is in series combination.

Let they are connected to the voltage V and the current in each resistance is I.

According to Ohm's law

Voltage across R', V' = I R'

Voltage across R'', V'' = I R''

Voltage across R''', V''' = I R'''

So, let the equivalent resistance is Rs.

I Rs = I R' + I R'' + I R'''

Rs = R' + R'' + R'''

(b)

When the three resistances R'. R'' and R''' is in parallel combination.

Let they are connected to the voltage V and the current in each resistance is I', I''. I'''.

Current in R', I' = V/R'

Current in R'', I'' = V/R''

Current in R''', I''' = V/R'''

The equivalent resistance is Rp.

V/Rp = V/R' + V/ R'' + V/R'''

1/Rp = 1/R' + 1/R'' + 1/R'''

6 0

3 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .