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gayaneshka [121]
1 year ago
7

A roller coaster moving along its track rolls into a circular loop of radius r. In the loop, it is only affected by its initial

velocity, gravity, and the shape of the track. Let v denote the instantaneous speed and a denote the magnitude of the instantaneous acceleration of the roller coaster in the loop. Which of the following is true in the loop?
a. The roller coaster is not in uniform circular motion, but we still have a=v^2/r everywhere on the loop
b. The roller coaster is not in uniform circular motion, but the tangential acceleration is so small that we can approximate a by v^2/r everywhere on the loop
c. The roller coaster is in uniform circular motion
d. The roller coaster is not in uniform circular motion, and a=v^2/r is only applicable at the very top and bottom of the loop where the tangential acceleration vanishes
Physics
2 answers:
earnstyle [38]1 year ago
5 0

Answer:

ok sooooooooooooo i  sure its the 3rd one but u know i could be wrong

Explanation:

stay safe 2021

crimeas [40]1 year ago
3 0

Answer:

c. The roller coaster is in uniform circular motion

Explanation:

Since the loop is circular with radius r, and its instantaneous speed, v is always constant, and also, its centripetal acceleration, a' = v²/r.

Since the angular speed, ω = v/r does not change, the magnitude of its  tangential acceleration is zero although there is a change in its direction because the direction of its initial velocity changes. That is a" = rα and α = Δω/Δt since Δω = 0, α = 0 and a" = r(0) = 0

So, there is no tangential acceleration. Since there is no tangential acceleration, our instantaneous acceleration which is the vector sum of our centripetal acceleration and tangential acceleration is a = √(a'² + a"²) =  √(a'² + 0²) = √a'² = a' = v²/r

So, a is always v²/r.

Since the instantaneous acceleration is always (a = v²/r) constant, the motion is uniform. So, it is uniform circular motion.

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
1 year ago
A friend climbs an apple tree and drops a 0.22-kg apple from rest to you, standing 3.5 m below. When you catch the apple, you br
ella [17]

Answer: (A) velocity = 2.8m/s

(B) Average force = 1.9536 Newtons

Explanation:

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3 0
2 years ago
A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds
horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

F\Delta t = m\Delta v

where:

F is the force exerted on the car

\Delta t is the duration of the collision

m = 1400 kg is the mass of the car

\Delta  v=-27 m/s is the change in velocity of the car

We can re-write the equation as

F=\frac{m\Delta v}{\Delta t}

In the 1st collision, the time is 1.5 seconds, so the force is

F_1=\frac{(1400)(-27)}{1.5}=-25,200 N

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

F_2=\frac{(1400)(-27)}{2.2}=-17,182 N

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

4 0
2 years ago
Read 2 more answers
Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a
Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

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Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0
1 year ago
What do fuel cells, batteries and, solar cells have in common A.the all produce static electricity B. they are all sources of di
Vaselesa [24]
The correct answer is B :)
3 0
2 years ago
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