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jok3333 [9.3K]
3 years ago
12

Sharks also swim in waters off the coast of Nova Scotia. When gray seals and harbor seals see sharks nearby, they must swim away

quickly or get eaten by the shark. Since the sharks and seals live in the same waters. could this interaction be called symbiosis? Why or why not? If not, what is this interaction called
Physics
1 answer:
Vlada [557]3 years ago
6 0
No. The seals are responding to a stimulus. 
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What family does the fox come from?
Nady [450]
Foxes belong to the dog family.
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3 years ago
You are riding in a hot air balloon that, relative to the ground, has a velocity of 6.0 m/s due east. You see a hawk moving dire
charle [14.2K]

Answer:Velocity = 6.325m/s

Directional angle= 18.43°

Explanation:

Using Right angle triangle

Let Velocity of ballon&hawk be VHB represent the height of the triangle.

Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.

Let Velocity of hawk and ground BE VHG represent the hypothesis.

Theta = opp/Adj= VHB/VBG

using pythagorean

VHG= SQRT(VHB^2+VBG^2)

VHG= sqrt(2^2+6^2)

VHG= sqrt(4+36)

VHG= 6.325m/s

Tan theta= 2/6

Tan theta =0.3333

Tan^-1 0.3333=18.43°

8 0
3 years ago
Does a comets tail always trail along behind it in its orbit?
Marat540 [252]
No, it only does when entering an atmosphere
4 0
3 years ago
The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

3 0
3 years ago
On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori
Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F  = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

                          W = F · S

Substituting the given values,

                           W = 89 N x 19 m

                           W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0
4 years ago
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