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stiv31 [10]
3 years ago
6

Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .

The river has a current that runs at a speed of 1.00 m/s. Note that speed is the magnitude of the velocity vector. The velocity vector tells you both how fast something is moving and in which direction it is moving. If you are swimming upstream (i.a., against the current), at what speed does your friend on the shore see you moving?
Physics
1 answer:
aliya0001 [1]3 years ago
6 0

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

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Answer:

D.

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An empty water tank has a rectangular base with side lengths of 1.2 m and 2.3 m. The weight of the tank applies a pressure of 35
LekaFEV [45]

Answer:

Weight = 966 Newton.

Explanation:

Given the following data;

Length = 1.2 m

Width = 2.3 m

Pressure = 350 Pa

To find the weight of the tank;

We know that weight is the force of gravity acting on an object multiplied by its mass.

Weight = mg = force

Hence, we would determine the force using the parameters that were given.

But we would first determine the area of the rectangular tank.

Area of rectangle, A = length * width

A = 1.2 * 2.3

A = 2.76 m²

Mathematically, pressure is given by the formula;

Pressure = force/area

Force = pressure * area

Substituting into the formula, we have;

Force = 2.76 * 350

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Therefore, the weight of the tank is 966 Newton.

5 0
3 years ago
What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J
Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)

and on the surface the current is

I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A

(b)

For Jb = J0(1 - r/R)

I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )

and on the surface

I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi  \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A

and in a simmilar way for Jb

I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi   J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

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Hyperopia: farsightedness

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Presbyopia occurs in old age people.

Explanation:

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Emmetropia: It is a vision without any defect.

Accommodation: It is the ability of the eye to adjust its focal length and adjusting the light on focus.

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