PHYSICS HIGH SCHOOL

Which refers to the rate of change in velocity?speed
O acceleration
O direction
O magnitude

Answers

Answer 1
Answer: The answer is obvious acceleration lol sorry if it sounds mean

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When a river or stream flows downward into a basin that is surrounded by high ground on all sides, what can form?

Answers

When a river or stream flows downward into a basin that is surrounded by high ground on all sides, a waterfall can form. The answer to the question above is 'waterfall' or 'waterfalls'. I hope this helps you on your assignments. You're welcome! 

Alvin and Theodore race across a parkinglot 360 cm long. Alvin travels at 22.03 cm/s
and Theodore at 17 cm/s.
When the Alvin crosses the finish line, how
far behind is Theodore?
Answer in units of cm.

Answers

Theodore will cross 82.2 cm behind Alvin

If you travel at 3 m/s for 12 seconds, how far did you travel?

Answers

Answer:

\boxed {\tt 36 \ meters }

Explanation:

We want to find how far was traveled, or the distance.

The formula for distance is:

d=s*t

where s is the speed and t is the time.

The speed is 3 meters per second and the time is 12 seconds.

s= 3 \ m/s \nt= 12 \ s

Substitute the values into the formula.

d= 3 \ m/s * 12 \ s

Multiply. Note that when multiply, the seconds, or s will cancel.

d= 3 \ m * 12

d= 36 \ m

d= 36 \ meters

The distance traveled was 36 meters.

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

According to Newton's first law, massive objects have _____ inertia than small objects, which means it takes more force to move bigger things than smaller ones.

Answers

According to Newton's first law, massive objects have more inertia than small objects, which Means it takes more force to move bigger things than smaller ones.

According to Newton's first law, massive objects have more inertia than small objects, which means it takes more force to move bigger things than smaller ones.

What does Krakauer hope to avoid by moving quickly up the face on the one rope? A) Altitude sickness. B) Frostbite. C) Falling into a crevasse. D) Running out of supplies.

Answers

By moving quickly up the face on one rope, Krakauer hopes to avoid falling into a crevasse (C).

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 electronq=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}.
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