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Roman55 [17]
3 years ago
13

What is the difference between the equilibrium vector and the resultant vector

Physics
1 answer:
Tresset [83]3 years ago
8 0

the difference between a resultant and equilibrant vector is that resultant vector is a direct quantity, one with both magnitude and direction, while the equilibrant vector is a force equal to, but opposite of, the resultant sum of vector forces, that force which balances other forces.

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Compare animal and plant cell
vagabundo [1.1K]

Answer:

A plant cell contains a large, singular vacuole that is used for storage and maintaining the shape of the cell. In contrast, animal cells have many, smaller vacuoles. Plant cells have a cell wall, as well as a cell membrane. Animal cells simply have a cell membrane, but no cell wall.

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2 years ago
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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

5 0
2 years ago
A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5
pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

v_w=v_s+v=1+1.5=2.5m/s

Distance=35 m

Time=\frac{distance}{speed}

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving=\frac{35}{v_w}=\frac{35}{2.5}=14s

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground=v_w=v_s+v=1-1.5=-0.5m/s

Time=\frac{-35}{-0.5}=70 s

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

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