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3 years ago

To crossing over the flooded canal.

Physics

1 answer:

zubka84 [21]3 years ago

4 0

Answer:

F. jumping

Explanation:

you can't throw/toss yourself, you cant roll over water, catching?, you cant run over water, jumps are bigger than hops

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Which structure is represented by the letter D? Choose 1 answer: (Choice A) A Chloroplast (Choice B) B Vacuole (Choice C) C Mito

Otrada [13]

Answer:

vacuole

Explanation:

Remember vacuoles in plant cells are a lot bigger than there animal cell counterparts.

6 0

3 years ago

What causes the different colors of visible light in the electromagnetic spectrum?

liraira [26]

Answer:

The color of the light is determined by the frequency of the light wave. Red, is lowest, frequency and violet is the highest.

5 0

4 years ago

You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s

BlackZzzverrR [31]

1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>

6 0

4 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

A small aircraft accelerated down a runway at 4.0 m/s²

Radda [10]

Given data in the problem :-

Acceleration (a) = 4.0 m/s^2
Initial velocity (u) = 0 m/s
Final velocity (v) = 34 m/s
Distance travelled by aircraft (S) = ?

From Newton's Laws of Motion we know that ,

v = u + at [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴ t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴ S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .

4 0

2 years ago