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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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Because a bond is not very easily forgotten if you work well with the person then you will work or
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Explanation:

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2 years ago
So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
shutvik [7]

Answer:

Answer is explained in the explanation section.

Explanation:

A)

Solution:

For this to find, we need to calculate the centripetal acceleration on the equator.

The centripetal acceleration of the equator:

a = 4\pi ^{2}RcosФ/T^{2},

where,

R is the radius of the earth

R = 6378 KM = 6.3 x 10^{6} and

T is the time period

T = 24 h = 86164.1 s

At Equator, Ф = 0°

So, CosФ = 1

Hence,

a = 4\pi ^{2}R/T^{2}

By plugging in the values, we get:

a = 4 x (3.14^{2}) x (6.3 x 10^{6}) / 86164.1^{2}

a = 0.03 m/s^{2}

Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/s^{2} which is very higher than the centripetal acceleration on the equator.

B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.

C) Of course, this is not a lie, it is true because the acceleration due to gravity is 9.8 m/ s^{2} and as we calculated the acceleration on the equator is 0.03 m/s^{2} which way too low to experience.

For percentage difference,

9.8 - 0.03 = 9.77

So, % diff = (9.8 - 9.77)/9.8 x 100

% diff = 0.00306 x 100

% diff = 0.306%

Obviously, this is way too low to experience.

D) With the help of same formula as discussed above, we have:

a = 4\pi ^{2}RcosФ/T^{2},

Here, Ф = 44.4°

Just putting the values. we get

a = 0.0241 m/s^{2}

Acceleration while sitting in my chair.

6 0
2 years ago
The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During th
Maksim231197 [3]

Answer:

17280 J or 17.28 kJ

Explanation:

Given that the voltage drop,

U = U2 - U1

U = 9 - 6

U = 3V

Also, we're told that the current, I is equal to 20 mA with the discharge time, t being 80 hrs.

Converting the time from h oi urs to seconds, we have

t = 80 * 3600

t = 288000

Now, to find the energy needed, we're going to use the formula

w = pt, where p = U * I

p = 3 * 20*10^-3

p = 60*10^-3

w = 60*10^-3 * 288000

w = 17280 J or 17.28 kJ

Therefore, the total energy the battery delivers in the 80 hrs is 17.28 kJ

5 0
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