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3 years ago

14

Some species of moths have "developed a unique defence" characteristic of sounding like a bat and therefore fooling a bats sense

of echolocation. What is this defence characteristic an example of?

1 answer:

Hunter-Best [27]3 years ago

6 0

I’m not smart sorry

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A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0

3 years ago

A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the fl

fiasKO [112]

Answer:

a

$KE = 7.17 *10^{7} \ J$

b

$t = 6411.09 \ s$

Explanation:

From the question we are told that

The radius of the flywheel is $r = 1.50 \ m$

The mass of the flywheel is $m = 430 \ kg$

The rotational speed of the flywheel is $w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec$

The power supplied by the motor is $P = 15.0 hp = 15 * 746 = 11190 \ W$

Generally the moment of inertia of the flywheel is mathematically represented as

$I = \frac{1}{2} mr^2$

substituting values

$I = \frac{1}{2} ( 430)(1.50)^2$

$I = 483.75 \ kgm^2$

The kinetic energy that is been stored is

$KE = \frac{1}{2} * I * w^2$

substituting values

$KE = \frac{1}{2} * 483.75 * (544.61)^2$

$KE = 7.17 *10^{7} \ J$

Generally power is mathematically represented as

$P = \frac{KE}{t}$

=> $t = \frac{KE}{P}$

substituting the value

$t = \frac{7.17 *10^{7}}{11190}$

$t = 6411.09 \ s$

3 0

3 years ago

Consider the hydrogen atom as described by the Bohr model. The nucleus of the hydrogen atom is a single proton. The electron rot

riadik2000 [5.3K]

To solve this problem we will apply the concept of centripetal acceleration. This type of acceleration is described as the product between the square of the angular velocity and the turning radius. Mathematically the expression can be expressed as

$a_c = \omega^2 r$

Here,

$\omega =$ Angular velocity

r = Radius

Our values are given as,

$\omega = 4.12*10^{16}rad/s$

$r = 5.29*10^{-11}$

Replacing,

$a_c = (4.12*10^{16})^2( 5.29*10^{-11})$

$a_c = 8.979*10^{22}m/s^2$

Therefore the electron's centripetal acceleration is $8.979*10^{22}m/s^2$

7 0

3 years ago

Why is it difficult to measure the volume of gas

MrRissso [65]

It's difficult to measure that because it's hard to make sure it is only a uniform layer of gas in whatever you're measuring it in

4 0

4 years ago

What will be the weight (acceleration due to gravity on the moon is 1/6 that of<br> the earth

jeyben [28]

Answer:

The weight will be 1/6 of whatever it is on Earth.

Explanation:

Remember F=ma

If the acceleration decreases by a factor of 6 while the mass stays constant, you can see the force (the weight) will also decrease by the same factor.

7 0

3 years ago