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Y_Kistochka [10]
3 years ago
6

What is the average acceleration of the particle between 0 seconds and 4 seconds? A. 0 meters/second2 B. 0.04 meters/second2 C.

0.06 meters/second2 D. 0.25 meters/second2 E. 0.928 meters/second2
Physics
1 answer:
lisov135 [29]3 years ago
6 0

Answer

D. 0.25 meters/second2

Explanation

The average acceleration is the ratio of change in velocity to the change in time of travel.Taking in this case that the change of velocity is a unit, then Average acceleration is given by;

Aacc=Vf-Vi/Tf-Ti

where Vf=final velocity,Vi=initial velocity' Tf=final time, Ti=initial time

Vf-Vi=1m/s

Tf-Ti=4-0=4seconds

Avacc=1/4=0.25m/s2

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Describe the ratio of kinetic energy and potential energy the skateboarder will have at each of the following points in her ride
amm1812

Answer:

no answer

Explanation:

7 0
3 years ago
Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19
aleksandr82 [10.1K]

Answer:

The answer to your question is distance between these electrons

                                                   = 1.386 x 10⁻¹⁴ m

Explanation:

Data

Force = F = 1.2 N

distance = d = ?

charge = q₁ = q₂ = 1.602 x 10⁻¹⁹ C

K = 8.987 x 10⁹ Nm²/C²

Formula

-To solve this problem use the Coulomb's equation

  F = kq₁q₂ / r²

-Solve for r²

  r² = kq₁q₂ / F

-Substitution

  r² = (8.987 x 10⁹)(1.602 x 10⁻¹⁹)(1.602 x 10⁻¹⁹) / 1.2

- Simplification

  r² = 2.306 x 10⁻²⁸ / 1.2

  r² = 1.922 x 10⁻²⁸

-Result

  r = 1.386 x 10⁻¹⁴ m

7 0
4 years ago
You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro
ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N

The spring obeys Hook's law:

F=k\Delta x

where k is the spring constant and \Delta x is the stretching of the spring. Since we know \Delta x=2.92 cm=0.0292 m, we can re-arrange the equation to find the spring constant:

k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance \Delta x is equal to the elastic potential energy stored by the spring, given by:

W=U=\frac{1}{2}k\Delta x^2

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

5 0
3 years ago
A swimmer swims 1000 m in the pool in 8.6 minutes. What was the average speed of the swimmer in m/s?
lawyer [7]

Speed = (distance covered) / (time to cover the distance)

Speed = (1,000 meters / 8.6 minutes) x (1 minute / 60 seconds) = 1.94 m/s
                                                                                                       (rounded)
6 0
3 years ago
Read 2 more answers
What about electrons allow them to be some of the fastest traveling subatomic particles?
rewona [7]
They're extremely small, occupying a very small volume, to the point where something like wind resistance that we think about with accelerating large objects like planes becomes completely irrelevant. A rogue electron can fly straight through most solid objects through the "empty space" between atoms. Their mass is also extremely small, 9.1*10⁻³¹ kg, making them relatively easy to accelerate to near light speeds (in comparison to other forms of matter) as it takes very little energy to set them into motion. Particle accelerators accelerate electrons to 99% of the speed of light in the real world every day.
7 0
4 years ago
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