Please help! This is due in 10 minutes!! I will mark brainliest asap

Answers

Answer 1

Answer: The object is at rest. This is because time is going by and there’s no distance being added

Answer 2

Answer: Third option is the answer aka the object is resting hope this helps you

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Answer to number 8 please
Um I don't remember how to do this, help :(

In which of the following are action and reaction forces involved?A. when hitting a baseball with a bat
B. walking
C. swimming
D. all of the above

Answers

Answer:

D. All of the above

Explanation:

Newton's third law states that:

"when an object A exerts a force on an object B, object B exerts an equal and opposite force on object A. These forces are called action and reaction"

Let's analyze each situation:

A. when hitting a baseball with a bat --> we have two objects (the baseball and the bat), so we have a pair of forces: the action is the force exerted by the baseball on the bat, while the reaction is the force exerted by the bat on the baseball

B. walking --> in this case, the two objects we should consider are the man walking and the road: the action is the force exerted by the feet of the man on the road, while the reaction is the force exerted by the road on the feet of the man, which allows him walking

C. swimming --> in this case, the two objects we should consider are the man swimming and the water around him: the action is the force exerted by the body of the man on the water, while the reaction is the force exerted by the water on the body of the man, which allows him to swim.

So, all the three choices are correct.

The use of oil or grease to lessen friction is called

Answers

They are called lubricants.

A space vehicle deploys its re-entry parachute when it's traveling at a vertical velocity of -150 meters/second (negative because the parachute is going down.) It comes to rest at 0 meters/second after 30 seconds. What's the average acceleration of the shuttle during this time span?

Answers

Acceleration, in physics, is the rate of change of velocity of an object with respect to time. An object's acceleration is the net result of any and all forces acting on the object, as described by Newton's Second Law.
a = change in velocity / change in timea = 0 - (-150) / 30 - 0a = 5 m/s2

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.5 x 10⁻⁷ C/m², and the plates are separated by a distance of 1.7 x 10⁻² m. How fast is the electron moving just before it reaches the positive plate?

Answers

The speed of the electron before reaching the positive plate is $1.30 * 10^(7)\ m / s$

Explanation:

As per Gauss law of electro statistics, the electric field generated by a capacitor is directly proportional to the surface charge density of the plate and inversely proportional to the dielectric constant. In simple words, the electric field is proportional to the surface charge density. So,

$\text {Electric field}=(\sigma)/(\varepsilon_(0))$

And then from the second law of motion, $F=m * acceleration$

So acceleration exerted by the electrons will be directly proportional to the force exerted on them and inversely proportional to the mass of the electron.

$Acceleration =(F)/(m)$

Since force is also calculated as product of charge with electric field in electrostatic force,

$\text {Acceleration}=(q E)/(m)=(q \sigma)/(m \varepsilon_(0))$

So, the charge of electron $q=1.6 * 10^(-19)\ \mathrm{C}, \sigma=\text { Charge per unit area }=2.5 * 10^(-7)\ \mathrm{C} / \mathrm{m}^(2)$

m is the mass of electron which is equal to $9.11 * 10^(-31)\ \mathrm{kg}$

$\varepsilon_(0)=8.85 * 10^(-12)\ \mathrm{Nm}^(2) \mathrm{C}^(-2)$

Then,

$\text { Acceleration }=(1.6 * 2.5 * 10^(-19) * 10^(-7))/(9.11 * 8.85 * 10^(-31) * 10^(-12))=(4 * 10^(-19-7))/(80.62 * 10^(-31-12))$

$\text { Acceleration }=0.0496 * 10^(-19-7+31+12)=0.0496 * 10^(17)\ \mathrm{m} / \mathrm{s}^(2)$

So the acceleration of the electron in the capacitor will be $4.96 * 10^(15) m / s^(2)$

Then, the velocity can be observed from the third equation of motion.

$v^(2)=u^(2)+2 a s$

As u = 0 and s is the distance of separation between two plates.

$\begin{array}{c}v^(2)=0+\left(2 * 4.96 * 10^(15) * 1.7 * 10^(-2)\right) \nv^(2)=16.864 * 10^(15-2)=16.864 * 10^(13)=1.684 * 10^(14)\end{array}$

Thus, $v=\sqrt{\left(1.68 * 10^(14)\right)}=1.30 * 10^(7)\ m/s$

So, the speed of the electron before reaching the positive plate is $1.30 * 10^(7) \mathrm{m} / \mathrm{s}$ .

Imagine you are holding an apple. Does this apple have energy?how do you know?

Answers

-- The apple has heat energy. If its temperature were Absolute Zero,
then the skin on my hand would freeze instantly and fall off.

-- The apple has potential energy, relative to the ground. If I were to
release it out of my hand, gravity would do work on it and pull it with
some speed to the ground, where it would have the kinetic energy to
break some blades of grass and dig a small crater in the soil.

-- The apple has chemical energy which, if I were to expose it to
human digestion and metabolism by eating it, would change to
glucose, soak into my muscles, and make itself available to me
whenever I need vim and vigor, mental alertness, visual acuity,
and the wherewithal to to escape any unexpected beast that might
in turn set upon me with the intention of converting MY entire person
into food energy for its own use.

The formula for speed is v = s/t.
a. True
b. False

Answers

if you mean S is the distance then it is true
Velocity = Distance / time

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