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3 years ago

Hey guys.. ans me

Physics

1 answer:

Stels [109]3 years ago

7 0

The time taken by the swimmer was 1 hour.

Why?

Since the swimmer is maintaining an angle of 150° while he was swimming, there were two components of the speed (horizontal and vertical). If we want to calculate the time taken by him to cross the river, we need to calculate the vertical speed and consider that the flow's speed is compensated by his horizontal speed.

We can calculate both components of the speed using the following formula:

$HorizontalSpeed=Speed*Cos(150\°)\\\\VerticalSpeed=Speed*Sin(150\°)$

Now, calculating we have:

$HorizontalSpeed=2\frac{Km}{h} *Cos(150\°)=-\sqrt{3} \frac{Km}{h} \\\\VerticalSpeed=2\frac{Km}{h} *Sin(150\°)=1\frac{Km}{h}$

Therefore, we have that the horizontal speed is compesating the flow's speed while his vertical speed is used to cross the river which is 1 Km wide.

Hence, we have that the tame taken is:

$Time=\frac{RiverWidth}{VerticalSpeed}=\frac{1Km}{1\frac{Km}{h} } =1h$

Have a nice day!

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A block whose weight is 45 N rests on a horizontal table. A horizontal force of 36 N is applied to the block. The coefficient of

riadik2000 [5.3K]

Answer:

1.5 m/s²

Explanation:

For the block to move, it must first overcome the static friction.

Fs = N μs

Fs = (45 N) (0.42)

Fs = 18.9 N

This is less than the 36 N applied, so the block will move. Since the block is moving, kinetic friction takes over. To find the block's acceleration, use Newton's second law:

∑F = ma

F − N μk = ma

36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a

6.75 N = 4.59 kg a

a = 1.47 m/s²

Rounded to two significant figures, the block's acceleration is 1.5 m/s².

Usually the coefficient of static friction is greater than the coefficient of kinetic friction. You might want to double check the problem statement, just to be sure.

6 0

3 years ago

Predict the energy transformation that occurs when carson sands a piece of wood with sandpaper.

tigry1 [53]

Answer:

Explanation:

The hand provides Kinetic Energy in moving.

The KE is transformed to Frictional energy

The Frictional Energy can produce heat and light energy.

The sandpaper produces little shreds from the grit of the paper.

The shreds have KE (they move)

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2 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

Is Smoke Abiotic or Biotic?

nlexa [21]

Answer: Smoke is Abiotic because it is not a living thing.

Explanation: Hopefully this helps u. Have a great rest of your day. I hope this is the right answer

6 0

3 years ago

A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of

Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0

3 years ago