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Vlada [557]
2 years ago
6

How do the echoes of the sounds of animals help them?

Physics
2 answers:
anastassius [24]2 years ago
7 0
Animals like bats and whales use something called echolocation (sounds or echoes that travel to the other animal) these echoes can let one animal let another one know that danger is near or that food is nearby etc it's like a telephone line
andre [41]2 years ago
4 0

Answer:

Echolocating animals emit calls out to the environment and listen to the echoes of those calls that return from various objects near them. They use these echoes to locate and identify the objects. Echolocation is used for navigation, foraging, and hunting in various environments.

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PLEASE HELPPPPP!!Calculate the net force acting on obiect A. In your answer, be sure to include the number, unit, and
asambeis [7]

Answer:

4 N to the right

Explanation:

For the object in this problem, we have 2 forces acting on it:

- A force of 2 N, to the right

- A force of 2 N, to the right

The two forces are in line: this means that the two vectors can be simply added by calculating their algebraic sum.

If we take right as positive direction, the two forces can be written as

F1 = +2 N

F2 = +2 N

So, the net force will be

F=F_1+F_2=+2+2=+4 N

And the positive sign means the direction is to the right.

6 0
3 years ago
Why is the gravitational potential energy of an object 1 meter above Earth's surface more than its potential energy 1 meter abov
Umnica [9.8K]
I Believe The Answer Is C
3 0
2 years ago
What situation would give you a mechanical advantage? *
const2013 [10]
D. Using a fixed-pulley system
3 0
2 years ago
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
2 years ago
A high jumper jumps 2.04 m. If the jumper has a mass of 67 kg, what is his gravitational potential energy at the highest point i
Mariulka [41]

Answer: 1339.5 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the jumper as he moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 67kg

g = 9.8m/s^2

h = 2.04 metres

Thus, GPE = 67kg x 9.8m/s^2 x 2.04m

GPE = 1339.5 joules

Thus, the gravitational potential energy at the highest point is 1339.5 joules

3 0
3 years ago
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