How do I do this? Help please!

A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative

navik [9.2K]

In this problem, we are given with the resultant velocity of the swimmer considering the current is running 0.91 m/s that is equal to 1.1 m/s2. In this case, using v = d/t, t = 3000m / 1.1 m/s equal to 2727.27 seconds equal to 45.45 hours. The distance downstream is equal to 2727.27 seconds * 0.91 m/s equal to 2481.82 meters.

3 0

3 years ago

Which phrases apply to the formation of the Grand Canyon?

Damm [24]

Answer:

Large scale

Explanation:

short term

6 0

3 years ago

A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

4 0

4 years ago

Lunar missions have revealed that the moon has:

myrzilka [38]

That the moon has soil within its shadowy craters rich and useful material

5 0

3 years ago

Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field

Scilla [17]

Answer:

Electric field lines point away from positive charges and toward negative charges. True

Electric field lines are continuous; they do not have a beginning or an ending. False

Electric field lines can never intersect. True

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. False

At every point in space, the electric field vector at that point is tangent to the electric field line through that point. True

Explanation:

Electric field lines point away from positive charges and toward negative charges. Always the field lines go to negative charges and leave from positive charges.

Electric field lines are continuous; they do not have a beginning or an ending. False

Because the field lines starts at positive charges and ends in negative charges.

Electric field lines can never intersect. True

It can not intercept the field lines because in that point the the field would have two directions. Besides, in that point the real value of the field should be found adding both field lines.

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. False

This fact is opposite to that so the regions of space where the magnitude the electric field is weak the lines are father apart and where the field is strong the lines are close together.

At every point in space, the electric field vector at that point is tangent to the electric field line through that point. True

This statement correspond to the definition of the field line.

5 0

3 years ago