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Fofino [41]
3 years ago
11

A force in the +x-direction has magnitude F = b/xn, where b and n are constants. For n > 1, calculate the work done on a part

icle by this force when the particle moves along the x-axis from x = x0 to infinity.
Physics
1 answer:
In-s [12.5K]3 years ago
3 0

Answer:

W=-\dfrac{b}{n-1}x_o^{1-n}

Explanation:

The magnitude of force in the +x direction is given by :

F=\dfrac{b}{x^n}

We need to find the work done. The integral form of work done is given by :

W=\int\limits {F.dx}

W=\int\limits^{\infty}_{x_{o}}{\dfrac{b}{x^n}.dx}

W=b \int\limits^{\infty}_{x_{o}}{\dfrac{1}{x^n}.dx}

W=b \int\limits^{\infty}_{x_{o}}{x^{-n}.dx}

W=\dfrac{bx^{1-n}}{1-n}|_{x_{o}}^\infty

W=-\dfrac{b}{n-1}x_o^{1-n}

Hence, this is the required solution.

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A cat starts to walk straight across a 100-meter-long field to her favorite tree. After 20 m, a dog sees the cat and chases it a
AleksAgata [21]

Answer:

8 seconds, Answer choice C.

Explanation:

The information they give us about the speed of the cat, is from the point at which the dog started chasing it (that velocity being 10 m/s).

Notice that the actual distance the cat run is: 100 meters minus 20 meters (100 - 20 = 80 meters). Therefore, we have information on the distance covered by the cat (80 meters), and its speed (10 m/s), so we can use the definition of speed to find the time it took the cat to get to the tree:

speed=\frac{distance}{time} \\10 \frac{m}{s} = \frac{80\,m}{time}\\time=\frac{80}{10} s=8 \,s

Since all units for the physical quantities involved were given in the SI system,  the answer comes also in the SI units of time: "seconds"

5 0
3 years ago
The following arbitrary measurements are made and the errors sited are the aximum errors A = 15.21 +0.01, B = 10.82 +0.05, C = 1
Molodets [167]

Maximum error in the result of the sum of measurement is equal to the sum absolute error of the individual observed measurements

(a) The maximum error in D is 0.09

(b) The standard error in D is approximately 0.034

The procedure for arriving at the above values is as follows;

The given measurements and the sited errors are;

A = 15.21 + 0.01

B = 10.82 + 0.05

C = 11.00 + 0.03

D = A + B + C

(a) Required parameter;

To calculate the maximum error in D

The equation for the propagation of error in addition is presented as follows;

Given that we have;

x = a + b

Therefore;

x + ±Δx = (a ± Δa) + (b ± Δb) = a + b ± (Δa + Δb)

∴ Δx = Δa + Δb

From the above formula, we have;

Where;

D = A + B + C

The maximum error in D = The sum of the maximum error in A, B, C

∴ The maximum error in D = 0.01 + 0.05 + 0.03 = 0.09

(b) Required parameter:

To find the standard error in D

The standard error is the sampling distribution's standard deviation, SD

Variance = SD²

The combined variance, SD² = The sum of the squares of individual standard deviations

Given that the standard errors represents the standard deviation, we get;

The combined variance, SD² = 0.01² + 0.05² + 0.03²

The combined variance, SD = √(0.01² + 0.05² + 0.03²) = 0.059

Standard \ error = \dfrac{SD}{\sqrt{n} }

Where n = 3, for the three measurement, we get;

Standard \ error = \dfrac{\sqrt{0.01^2 + 0.05^2 + 0.03^2} }{\sqrt{3} }  \approx 0.034

The standard error in D is approximately 0.034

Learn more about maximum error and standard error here:

brainly.com/question/13106593

brainly.com/question/17164235

3 0
3 years ago
Three point charges are placed at distances of d , 2 d , and 3 d from a point P. The particle that is 2 d away from P has a char
mel-nik [20]

Answer:

at d the charge will be 3q and at 3d it will be 9q

Explanation:

for V=Vp-V2d

V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q

7 0
3 years ago
The nucleus of a cell is like the ____ of a large animal.
djverab [1.8K]
It is either the brain or the stomach.
3 0
3 years ago
A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop
KengaRu [80]

#A

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

\\ \rm\longmapsto Force=ma

\\ \rm\longmapsto Force=4.4(-1.74)

\\ \rm\longmapsto Force=-7.656N

#B

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

\\ \rm\longmapsto a=\dfrac{v-u}{t}

\\ \rm\longmapsto u=v-at

\\ \rm\longmapsto u=0-(-1.74)(1.27)

\\ \rm\longmapsto u=1.74(1.27)

\\ \rm\longmapsto u=2.2m/s

4 0
3 years ago
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