PHYSICS HIGH SCHOOL

A car moves at 12 m/s for 30.0 seconds. It then accelerates at 1.5 m/s2 for 8.00 seconds. Finally, it continues on at this top speed for another 12.0 seconds. Calculate the net
displacement during the whole time interval.

Answers

Answer 1
Answer:

Over the first 30.0 s, the undergoes a displacement of

(12 m/s) * (30.0 s) = 360 m

Over the next 8.00 s, the car accelerates from 12 m/s to a top speed of

12 m/s + (1.5 m/s²) * (8.00 s) = 24 m/s

and over this time interval, it is displaced an additional

(12 m/s) * (8.00 s) + 1/2 (1.5 m/s²) * (8.00 s)² = 144 m

For the last 12.0 s, the car moves at a constant speed of 24 m/s to cover a distance of

(24 m/s) * (12.0 s) = 288 m

So the car's net displacement is 360 m + 144 m + 288 m = 792 m. (The net displacement is the same as distance in this case because the car moves in only one direction.)

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An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uniform magnetic field ???? = 0.030T ???? − 0.15T ???? . (a) Find the force on the electron due to the magnetic field (b) Repeat your calculation for a proton having the same velocity.

Answers

Explanation:

It is given that,

Velocity of the electron, v=(2* 10^6i+3* 10^6j)\ m/s

Magnetic field, B=(0.030i-0.15j)\ T

Charge of electron, q_e=-1.6* 10^(-19)\ C

(a) Let F_e is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

F_e=q_e(v* B)

F_e=1.6* 10^(-19)* [(2* 10^6i+3* 10^6j)* (0.030i-0.15j)]

F_e=-1.6* 10^(-19)* (-390000)(k)

F_e=6.24* 10^(-14)k\ N

(b) The charge of electron, q_p=1.6* 10^(-19)\ C

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

Physics help!! 10ptA small object placed in the gravitational field of a more massive object will

A. accelerate toward the more massive object.
B. accelerate away from the more massive object.
C. move toward the more massive object at a constant speed.
D. move away from the more massive object at a constant speed.

Answers

A. accelerate toward the more massive object is your answer. In fact, even the acceleration with accelerate, also known as the jer.k. The more massive object will pull on the smaller object and the smaller object will increase its speed towards the more massive object.


idk why jer.k is a swear

Gravitational forces always appear in equal and opposite pairs.

When a small object is placed in a position near a more massive
object, there is a gravitational force on each object, attracting it
and drawing it toward the other one.

The force on the small object causes it to accelerate toward
the more massive one.  The force on the more massive object
causes it to accelerate toward the smaller one.

The forces are equal.  But since acceleration is inversely
proportional to mass, the small object heads toward the
more massive one with greater acceleration, and the more
massive object heads toward the smaller one with smaller
acceleration.

When you jump into the pool from the high diving board,
you accelerate toward the Earth, and the Earth accelerates
toward you.  The gravitational forces between you and the
Earth are equal in both directions.  But the Earth has more
mass than you have, so its acceleration toward you is smaller
than your acceleration toward it.

A particle with velocity v(t)=1/t​−3, in m/s, moves in a straight line. Its initial displacement is known to be s(1)=1, in m. Find the position function s(t).

Answers

Answer:

s(t) = (t1 - t2)* [((1/t1 - 3) + (1/t0 - 3))/2]

Explanation:

We will assume that v(t) is in units of m/s and t (time) is in seconds.

v(t)=1/t​−3

At time 0 (initial) the equation tells us the particle has a velocity of

v(t)=1/t​−3

v(0)=1/(0)​−3

v(0) = - 3 m/s

The particle is moving from right to left (the negative sign) at a rate of 3 m/s.

The position of the particle would be the average velocity times the time traveled.

Distance = Velocity x Time  (with Velocity being the average between times t0 and t1)

We'll use s(t) for displacement for time t.

 s(t) = v*t

We need the average velocity for the time period t0 to t1.

Let t0 and t1 be the initial and final times in which the measurement takes place.

At time t0 the velocity is = 1/t0 - 3

At time t1 the velocity is = 1/t1 - 3

The displacement is the average velocity between the two points, t0 and t1.  This can be written as:

 [(1/t1 - 3) + (1/t0 - 3)/\]/2

Displacement: s(t) = (t1 - t2)* [((1/t1 - 3) + (1/t0 - 3))/2]

   

The source of the centripetal force that arises when a runner rounds a curve on a track is _____.tension
friction
weight
gravity

Answers

while taking a turn on a track, the track provides the frictional force acting towards the center of the curve. we know that to move in a curve, centripetal force is required. here since the frictional force acts towards the center, it provides the necessary centripetal force to move in circle. hence the correct choice is

friction

It's actually Friction.

I just did the test and got it right.

Plates move apart at ____ boundaries.a. convergent
c. divergent
b. stable
d. transform

Answers

The correct answer is C. divergent. Divergent boundaries are something like lines found between two tectonic plates. These constructive boundaries, or extensional boundaries, are formed and exist between two tectonic plates that move away from each other. Convergent would thus be where they meet. This is all according to the tectonic theory.

Plates move apart at divergent boundaries, which occur when two tectonic plates are moving away from each other.

Plates move apart at divergent boundaries. Divergent boundaries occur when two tectonic plates are moving away from each other. This movement creates a gap or rift between the plates where molten rock rises to fill the space, forming new crust. The best example of a divergent boundary is the Mid-Atlantic Ridge.

Learn more about Tectonic Plates here:

brainly.com/question/34545297

#SPJ6

The half-life of carbon-14 is 5,730 years. How many grams of parent material will remain from a 120-gram sample of carbon-14 after 2 half-lives?

Answers

After 2 half lives, one half of one half of the original material remains. That's one fourth. One fourth of 120 grams is 30 grams remaining.
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