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3 years ago

Asexual reproduction provides an advantage to organisms under which of the following conditions?

Physics

1 answer:

krek1111 [17]3 years ago

4 0

Answer:

The organisms are widely dispersed in the habitat and rarely encounter each other...

Explanation:

As asexual reproduction needs only 1 parent.. they can do it on their own.. without encountering their mates

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Two planets have the same surface gravity, but planet b has twice the radius of planet

nevsk [136]

The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
m₂ = 4m

<em>Thus, the answer is D.</em>

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3 years ago

You are falling off the edge.what should you do to avoid falling..?

Nadya [2.5K]

Lean your shoulders back and your waist forwards. Use your arms as a counter weight.

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3 years ago

Read 2 more answers

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t

fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

$A \rightarrow (A-\frac{A}{9} )$

$A \rightarrow \frac{8A}{9}$

As we know,

Energy will be:

⇒ $E_{1}=\frac{1}{2}KA^2$

and,

⇒ $E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2$

$=\frac{64KA^2}{162}$

⇒ $\Delta E=E_1-E_2$

On putting the estimated values, we get

$=\frac{1}{2}KA^2-\frac{64KA^2}{162}$

⇒ $\frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}$

$=\frac{40}{162}$

$=0.246$

3 0

3 years ago

‏A 50 - N x m torque acts on a wheel with a moment of inertia 150 kg x m² . If the wheel starts from rest , how long will it tak

denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, $\triangle \theta = 2\pi$

Torque, $\tau = 50 Nm$

Moment of Inertia, $I = 150 kg m^2$

If the wheel starts from rest, $w_{0} = 0 rad/s$

The angular displacement of the wheel can be given by the formula:

$\triangle \theta = \omega_0 t + 0.5 \alpha t^2$ ................(1)

Where $\alpha$ is the angular acceleration

$\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2$

To get t, put all necessary parameters into equation (1)

$2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s$

3 0

3 years ago

What will be the value of both charges if they are 5 cm apart and suffer a

Mrac [35]

Answer:

$\boxed{q = 1.2 \times {10}^{ - 6} C}$

Explanation:

$f_e = \frac{{q}^{2}k }{ {r}^{2} } \\ q = \sqrt{ \frac{f_e( {r}^{2} )}{k} } = \sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } \\ q =\sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } = \sqrt{ \frac{0.013}{9 \times {10}^{9} } } \\ q = 1.2 \times {10}^{ - 6}$

3 0

2 years ago