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koban [17]
3 years ago
12

In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for

F⃗ ? Let U=0 when x=0.
Physics
1 answer:
Blizzard [7]3 years ago
7 0

Answer:

The potential energy is -4x^3

Explanation:

Given that,

Force F=-\alpha x^2 i

We need to calculate the potential energy

Using formula of work done

\Delta U=F(x) dx

Put the value of F

\Delta U=-\alpha x^2\ dx

On integration

U=-\alpha \dfrac{x^3}{3}+C

U=-\dfrac{\alpha x^3}{3}+C...(I)

U = 0, x = 0 so C = 0

Put the value of c and α in equation (I)

U=-\dfrac{12x^3}{3}+0

U=-4x^3

Hence, The potential energy is -4x^3

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Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown
jonny [76]

Answer:

Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle

Explanation:

When you centrifuge precipitate it enables the nucleation to form.

Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.

5 0
3 years ago
Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want
denpristay [2]

Answer:

2.32 s

Explanation:

Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of  the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

Given: s = 10 m, g' = 3.7 m/s²

Substitute this value into equation 2

t = √(2×10/3.7)

t = √(20/3.7)

t = √(5.405)

t = 2.32 s.

4 0
3 years ago
Read 2 more answers
You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed
Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

a=1.1 g

where g=9.8 m/s^2 is the acceleration of gravity. Substituting,

a=(1.1)(9.8)=10.8 m/s^2

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0
3 years ago
(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en
V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0
3 years ago
How is a scientific law alike as a societal law
bija089 [108]
Describe that a scientific law is a description of a specific relationship under given conditions. Describe that scientific laws are developed from large amounts of experimental observations that result in the same outcome.
5 0
3 years ago
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